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Brief User Guide for TI-83 Plus and TI-84 Business and Economics


INDEX:

To facilitate lookup, the instructions are divided into the following categories:
I.  Supply, Demand, and Costs -  Equilibrium point; Marginal average cost;  Marginal cost and “Next Item”
    cost; Maximum Profit;
Price, Revenue, and Quantity at given demand; determining price-demand function
   from given data,

II.  Miscellaneous - Harmonic mean, Geometric mean, Mean of grouped data, How long to double interest, Leontief
    input-Output problem,
III.  Linear Programming - Graphical Method; Graphing Using the Inequality APP, Simplex Method,
IV.  Calculus Applications - Cost, Average Cost, Marginal Cost, Average-Cost Minimization, Cost minimization,
     Complementary/Substitute Product Determination,
Equality of Marginal Product of Labor and Capital, Consumers' surplus,
     Suppliers' surplus

 

IRELEASE DATE:  2/2/10       DATE LAST REVISED:  2/10/12
© 2003 Frank Kizer   NOTE:  See copy restrictions and printing hints at the end of this document.  

General:  Some of these problems can be solved more easily with a little calculus and some manipulations with paper
    and pencil.  Nevertheless, I have included some rather simple problems as examples of how a graphing calculator
    can be used.

I.  Supply, Demand, Costs and Related Items:

    1. Equilibrium Point:
       
A certain produce has a demand function p = -2x² +80 and a supply function of p= 15x +25.  If x    
        represents the demand quantity in thousands and p the unit price in dollars, find the  
        equilibrium quantity and price.
        a)  Press Y= and enter the demand equation opposite Y1 and the supply equation opposite Y2 .     
             Use the [x, T, 0, n] key to enter “x” and the x² key to enter the exponent.
      b)  Press Zoom, 0 (ZoomFit) to get the graphs on the screen.  Now press WINDOW and set  
           Xmin  and Ymin =0 and press GRAPH.
     c)  Press 2ND, CALC, 5(Intersect).
     d)  From the graphing screen that appears, press ENTER, ENTER; then move the cursor  
         approximately to the intersection of the graphs and press ENTER again.
   e)  The equilibrium point price, y= 67.45, and equilibrium quantity, x=2.69699, will be displayed.  Multiply the x-value by
         1000 to get 2689.99 or 2690 rounded off. 

2.  Marginal Average Cost:
     The cost function for manufacturing a certain household item is c(x) = 5000+2x.  Find the   
      marginal average cost for 500 units.
     Preliminary:
  The average cost function is c(x)/c = (5000+2x)/x.
     a)  Enter (5000+2x)/x opposite Y1.
     b)  Set Xmin=0, Xmax=550, Ymin=0, and Ymax=150 and press GRAPH.
     c)  Press 2ND, CALC, 6(dy/dx).
    d)  Press ENTER and the value -0.02 will be displayed.

3.  Marginal Cost and “Next Item” Cost :
    
The cost of a certain product is given by c=700+0.03x +.0002x² .  Compare the marginal cost and  
     the “next item” cost at 150 units.
     a)  Press Y= and enter 700+0.03x +.0002x²  opposite Y1 .  Use the [x, T, 0, n] key to enter “x” and   
           the x² key to enter the exponent.  Press ZOOM, move the cursor to ZoomFit and press ENTER.
     b)  Press 2ND, CALC, 6(dy/dx). And press ENTER.
     c)  On the screen that appears, enter 150 and press ENTER to get dy/cy=.09 for the marginal   
           cost.
 
Now, find the cost of item 151.
     a)  Press 2ND, CALC, ENTER, for Value.
     b)  ENTER 150 and press ENTER to get 709.
     c)  Press 2ND, CALC, ENTER.
    d)  Enter 151 and press ENTER to get 709.902.
    e)  This gives 709.902-709=0.0902 for the actual cost of manufacturing item 151.

4.  Finding Maximum Profit:
    
The demand of a certain product is represented by p=40/√q and the cost of producing q items
     is given by c= .04x +400.  Find the number of items that gives maximum profit and the price  
     that yields maximum profit.
     We know, of course, that the maximum profit occurs when the cost of producing another item
      (approximated by the marginal cost)) is equal to the revenue and on that item (marginal
      revenue).
      P=R-C
        = xp(x) - C(x)
       
=x(40/√(x)) –(0.4x +400)
        =40√(x) –0.4x -400
     a)  Enter 40√(x) –0.4x -400 opposite Y1  using the [x,T,0,n] key to enter “x.”
     b)  Press WINDOW and set x-min=0, Xmax=5000, Ymin=300, and Ymax=700.  Press GRAPH.
     c)  Press 2ND, CALC, 4(maximum). 
    d)  Move the cursor somewhat to the left of the maximum of the curve and press ENTER.
             Move the cursor somewhat to the right of the maximum of the curve and press ENTER.
              Finally, move the cursor approximately to the peak and press ENTER again. 
     e)   The terms displayed will indicate x=2500 and y=600.  So, the maximum profit occurs at 2500
            items and the maximum profit is $600.
     f)    From the price equation, do the following calculation:
               p= 40/√x
               p= 40/√2500
               p=0.80

5.  Price at a Given Demand:

     The relation between price, p,  and quantity demanded, q, is given by the equation
      p² =12000-2x.  Find the price when the demand is 3900 and 4100.
    First solve for the price as a function of x.
      p² =12000-2x
      p=√(12000-2x)
      Ex 1:
      
a)   Press Y= and enter  √(12000-2x) opposite Y1.  Use the [x,T,0,n] key to enter “x.”
        b)  Press WINDOW and set Xmin=0, Xmax=6500, Ymin=0, and Ymax=110.  Press GRAPH to  
             display the graph.
       c)  Press 2ND, CALC, ENTER and enter 3900 opposite X.
       d)  Press ENTER and the answer 64.807… will be displayed.
       e)  After you have recorded the last  answer, enter 4100 and press ENTER to get 61.644…
       Ex 2:

       For the price-demand relation of Ex. 1 above, find the maximum revenue and the quantity  
       that gives the maximum revenue.

       a)  Enter  x√(12000-2x),  the expression for R(x), opposite Y2.
       b)  Move the cursor to the = sign at Y1 and press ENTER to disable the graph for   
            revenue.  Press ZOOM, 0(ZoomFit) to display the graph.
       c)  Press 2ND, CALC 4(maximum).
       d)  When the graph appears, move the cursor somewhat to the left of the peak and  
            press ENTER.  Move the cursor somewhat to the right of the peak and press
            ENTER.   Move the cursor approximately to the peak and press ENTER again.
       e)  A quantity of x=4000 and revenue of y=252982.21 will be displayed.
     Ex. 3: 

       Find the price at maximum revenue.
        a)  Disable Y2 by moving the cursor to the = sign and pressing ENTER.  Enable Y1 by moving the cursor
              to the = sign opposite Y1 and pressing ENTER.
        b)  Press ZOOM, 0(ZoomFit) to display the graph.
        c)  Press 2ND, CALC, ENTER.
        d)  From the graph screen, enter 4000 opposite x and press ENTER.  The price   
             63.245… will be displayed.

6.  Price-Demand Function from Given Data:
     I will give the method for doing this problem totally with the calculator; then I will give a method for doing
    detailed calculations to find the constants “a” and “b” in the best-fit equation

     A company determines the price-demand relationship of one of its products is related as given in the table
     below.  Determine the price-demand function,  the price at 8000 units, the maximum revenue and the value at
     at which maximum revenue occurs. 

Quantity (x 103 ) Price
1000 72
4000 63
9000 48
14000 33
20000 15

     a)  Press STAT, ENTER and enter the quantities in list L1 and the prices in L2.
      b)  Press STAT, move the cursor to CALC and press 4 (linReg (ax+b)). 
      c)  Press VARS, move the cursor to Y-VARS, and press ENTER, ENTER.   LinReg(a+bx)  Y1  should be displayed
           on the home screen.  Note that if the data are not in the default lists, L1 and L2, you will need to enter the lists
           separated by commas before doing the steps to enter Y1.
     d)  Press ENTER and the values for "a," -0.003, and "b, "  75, will be displayed along with other results. So, the
          demand function is P= -0.003X+75.
     Now, we want to determine the price for at 8000 units from the graph of the demand function.
      e)  Press Y= and observe that  -.003X+75 is displayed opposite Y1.
      f)   Press WINDOW and set Xmin=0, Xmax=8500, andYmin=0.  Press ZOOM, 0(ZoomFit), and the graph will be
           displayed.
      g)  Press 2ND, CALC, ENTER to implement the Value function.
      h)  Enter 8000 opposite "X" and press ENTER to obtain 51 for the price.
     Now, find the maximum revenue.
      a)  Press Y=, disable Y1 by moving the cursor  to the = sign opposite Y1 and pressing ENTER.
      b)   Opposite Y2, enter "x."  Use the [X,T,0,n] key to enter "x."
      c)  With the cursor opposite Y2 , press VARS, move the cursor to Y-VARS and press ENTER, ENTER.
           You should now have XY1 displayed opposite Y2.
      d)  Press WINDOW and set Xmin=0, Xmax = 25000, Ymin=0.  Press ZOOM, 0 (ZoomFit),  and the graph
            will be displayed.
       e)  Press 2ND, CALC, 4(maximum) and the graph will be again displayed.
       f)  From the graph screen, move the cursor somewhat to the left of the peak of the graph and press ENTER.
            Move the cursor to slightly to the right of the peak and press ENTER again.  Finally, move the cursor
            to the approximate peak of the graph and press ENTER again.  X=12500 and Y=468750 will be displayed. 
            These are the quantity at which maximum revenue occurs and the maximum revenue at that quantity.
      

Detailed Calculations:  

Since many teachers use the detailed calculations as an instructional  method, I will now go through a fairly detailed
calculation of the constants “a” and “b.”  Assume that we have the  x- and y-values as listed in the table in the first part
 of this problem.

L1 Quantity (x) (x 103 )

L2   Price (y)

 

L3   x²

L4 xy

1000

72

1E6

72000

4000

63

1.6E7

252000

9000

48

8.1E7

432000

14000

33

1.96E8

462000

20000

15

4E8

300000

xi = 48000

yi= 231

∑ xi² = 6.94E8

xiyi=1.52E6
 

 We will now calculate the values for x² and xy which have already been entered in the table on this page.
a) Press STAT, ENTER, ENTER to go to the list screen.  Enter the values in the table into the
 first two columns of the lists as indicated in the table above.
b)   Highlight the name, L3, and the top of the lists and press 2nd, L1, x², ENTER.  The squared
       numbers will be stored in list L3 .
c)  Highlight the name for list L4 and press 2nd, L1, 2nd, L2, ENTER.
Now, solve the system of equations indicated as follows:
 ∑x i)b +(∑xi²)m =  ∑xiyi
      
nb   +(∑x i)m =  ∑yi
 In these equations, “m” is the slope and “b” is the y-intercept.  Those may be designated by such constants
as "a" and "b" in some texts. 
In terms of our lists in the table above, these equations can be written as a matrix as follows:
∑L1 +(∑L3
=  ∑L4
   
n   +(∑L2 =  ∑L3
Enter the Coefficients into the Matrix:
       a)  Press 2nd, MATRIX, go to EDIT and press ENTER.
      b)  Create a 2x3 matrix by pressing 2, ENTER, 3, ENTER.  Then move the cursor to the first element to enter the data.
            We are now going to sum the lists and enter the sums in the matrix  that we just edited.
      c) Press 2nd, LIST, move the cursor to MATH, and press 5 to paste sum(  opposite 1,1 at the bottom of the screen.
      d)   Enter L1, ),  so that you have sum(L1).  Now press enter to get the sum in the first element of the matrix.  
      e)  Repeat this procedure for  L3 in element (1,2),  L4 in element (1,3); then enter the number for "n" in element (2,1). 
      f)  In a similar manner,  repeat the sums for L2 in element (2,2) and L3 in element (2,3).
                When you finish solve the matrix for "a" and "b" as follows: 
      f) Press 2nd, QUIT to get out of the matrix.  Then press 2nd, MATRIX, move the cursor over to MATH,  and then select rref( and press.
       e) Press 2nd, MATRIX and then ENTER if you have entered your numbers in [A], otherwise select the proper matrix
            and press ENTER.  You should now have rref([A] on the screen.
      f )  Press ENTER to get the answer.  The first row will be "b" and the second will be "a."  You'll get almost the same answer
           as with all of the arithmetic.
      COMMENT:  The matrix method may seem like a formidable number of steps, but is saves many, many individual arithmetic
      steps. 

II.  Miscellaneous:
      1. Harmonic mean:
          Suppose that an investor has $1000 to invest each month and he pays $8.00, $8.50, $9.00, $9.50, $10.00, and $10.50 for the
          for the first six months.  What is his average cost per share?
          a)  Press STAT, ENTER to go to the lists and enter the data in list L1.
          b) Place the cursor on the name for L2 and enter 1, ÷, 2nd, L1, so that you have 1/L1..  Press ENTER and the reciprocals will
              be displayed

        
c) Press 2nd, QUIT to leave the list editor and press 1, ÷, (.  Press 2nd, LIST, move the cursor to MATH, and press 5 for sum(.
         d) Press 2nd, L2, ), ÷, 6, ).  You should now have 1/(sum(L2)/6)
         e) Press ENTER and $9.1706....will be displayed as opposed to the arithmetic mean of $9.25.

     2. Geometric mean:
          Suppose that during the past five years you has 75%, 10%, 30%, 20%, and -60% returns on your investment.  What would be
          your average return for the five-year period?  You can enter ((1.75*1.1*1.3*1.2*.4)^(1/5))-1, but if you have a more complex
          problem or you don't like entering number, you can do it as follows:
          a)  Press STAT, ENTER and enter the percentages in list L1.
          b)  Move the cursor to highlight the name for L2 and press 2nd, L1, ÷, 100, +, 1.  So that you have L1/100 +1 and press ENTER.
          c) Move the cursor to the first blank space in L2 and press (,; l then press 2nd, LIST.  Move the cursor to MATH and press 6 to
              paste prod( to the bottom of the list screen. 
         d) Press 2nd, L2, ), ^, (1, ÷, 2nd, 5)-, 1.  You should now have (prod(L2))^(1/5) -1 at the bottom of the list screen. 
          e) Press ENTER and you will get .03734 or 3.734% rather than 15% for the arithmetic mean.

 3)  Median of Grouped data:
      
Consider the following table.

Age

5-14

15-24

25-34

35-44

45-54

Midpoint

9.5

19.5

29.5

39.5

49.5

Freq (People)

750

2005

1950

195

100

a)  To find the median class, enter the midpoints in L1 and the frequencies in L2 .
b)  Press STAT, move the cursor to highlight CALC and press ENTER.  1-Var Stats will be displayed on the home screen.
c)  If the data are in lists L1 and L2 just press ENTER.  If they are in other lists, you must enter the lists.  For example,  
     press 2ND, L2, comma, 2ND, L3 and then press ENTER
d)  Press ENTER and scroll down to Med=19.5.  That is the median of the class 15- 
     24. So 15-24 is the median class.
e)  Enter the appropriate data into the following formula:
Median = L + I *(N/2 - F)/f
Where
L = lower boundary of the interval containing the median.
I = width of the interval containing the median.
N = total number of respondents.
F = cumulative frequency of those below the median class.
f = number of cases in the median class.
f)  When you are finished entering, you should have this:
    14.5+10(5000/2-750)/2005
g)  Press ENTER and you should get 23.228… Notice that the answer is different form
     the value of 19.5 given by the calculator. That value of 19.5 was chosen by merely finding the  midpoint 
     of the median class.

 4.  How Long to Double Interest:
      Although this could be found using the Finance Application, here's a quick way to do it.
      An investment has a 7% interest compounded annually.  How long will it take for the value
      to double?
     Preliminaries:   From the equation A=P(1+r/n)nt , we have 2P=P(1+0.07)t .  Cancel "P" and we have
     2=1.07t , or 2=1.07x for the calculator.
     a)  Press Y= and enter 2 opposite Y1; then enter 1.07^x opposite Y2.
     b)  Press WINDOW and set Xmin=0, Xmax=15, Ymin=0 and Ymax=3. Press GRAPH.
     c)  Press 2ND, CALC, 5 (for Intersect).
     d)  From the graph screen that appears, press ENTER, ENTER; then move the cursor near the intersection
           of the two lines and press ENTER again.  The answer of 10.24 years will be displayed.
       Note that you can find other factors such as triple or quadruple by merely changing Y1 to 3 or 4.

5.  Leonfief Input-Output Problem:
    
Suppose that an economy consist of the sectors of food, clothing, and shelter.  In that economy, producing one unit of food
     requires 0.4 units of food, 0.2 unit of clothing, and 0.2 units of shelter Producing one unit of clothing requires 0.1 units of
    food, 0.2 units of clothing, and 0.3 units of shelter.  Producing one unit of shelter requires 0.3 units of food, 0.1 unit of clothing,
    and 0.1 unit of shelter.  The economic demands are $100-million of food, $30-million of clothing and $250-million of shelter. 
    Find the level of production needed to meet  these demands.
    We can write the following input-output matrix:
    

  F S C
F 0.4 0.2 0.2
C 0.1 0.2 0.3
S 0.3 0.1 0.1

And the following for the demand matrix:

100
30
250

From the Leontief input-output model we have this:
X-AX =D
X=(1-A)-1 D
a)  Press 2ND, MATRIX, move the cursor  to EDIT and enter the input-output matrix in matrix [A].  Press 2ND, QUIT to
      leave the matrix editor.
b)  Press 2ND, MATRIX, move the cursor  to EDIT and enter the demand matrix in matrix [B].  Press 2ND, QUIT to
      leave the matrix editor.
      Now, we want to enter the appropriate expression for (1-A)-1 D.
c)  Press 2ND, MATRIX, move the cursor to MATH and press 5(identity), 3, ), - (minus sign).
d)  Press 2ND, MATRIX, ENTER, ).  You should now have (identity(3)-[A]) on the home screen.
e)  Press x-1 , 2ND, MATRIX , 2(for matrix [B]), You should now have (identity(3)-[A])-1 [B] on the home screen.
f)  Press ENTER to get 396.34 for food, 252.22 for clothing, and 437.80 for shelter.  All are rounded to two decimal
    places.

III.  Linear Programming:
  
   1. Graphical Method:
          Ex 1: 
Suppose that we want to maximize the objective function with function and constraints defined
          as follows:
          z=2x+5y
          3x+2y≤6
             -x+2y≤ 4
           x
≥0, y≥0
         a)  First rewrite the inequalities in slope-intercept form:
             y≤(-3/2)x +3
                  y≤x/2+2 
         b)  Press Y= and enter the expressions in Y1 and Y2.  If you wish, you can do the shading by moving the
             cursor over to the left and pressing ENTER until you get the third-quadrant triangle for <, but I prefer to
            just graph the equations.
         c)  Press WINDOW and set the following: Xmin=0, Xmax=5, Ymin=0, Ymax=4.  Press GRAPH and the
              graphs will be displayed.  You can read the points (0,2) and (2,0) from the graph, but we want to find
              the point of intersection. 
         d)  To find the intersection point press 2ND, CALC, 5(Intersect), move the cursor back to about x=.7 and
              press ENTER. Press ENTER, ENTER  after the cursor has moved to the other curve and x=.5, y=2.25
              will be displayed.
          f)  We now have the points (0,2), (.5, 2.25), and (2,0) that we want to plug into the objective function. You
              could do this manually fairly easily, but there may be occasions when it is not done so easily by hand. 
              Do the calculations with the calculator as follows:
              1)  Press 0, STO, X (use the [x,T,0,n] button for X), ALPHA, : (the decimal button), 2, STO, ALPHA, Y,
                  ALPHA, :,  2, X, +,  5, ALPHA, Y. You should now have 0
→X:2→Y:2X+5Y.
                   2)  Press ENTER and 10 will be diplayed.
                   3)  Now, press 2ND, ENTER and change the stored values of X and Y, so that you have the following:
                         .5→X:2.5→Y:2X+5Y.  Press ENTER and 13.5 will be displayed.
                    3)  Now, press 2ND, ENTER and change the stored values of X and Y, so that you have the following:
                         2→X:0→Y:2X+5Y.  Press ENTER and 4 will be displayed.
                   So, the point (.5, 2.25) maximizes the objective function. 
               Using the Inequalz APP:
      
2)  Graphical Linear Programming with the Inequalities Application:
             
Do the same problem as above, that is, find the maximum of the objective function z=2x +5y, subject to the following constraints:
             
 3x+2y≤6
              -x+2y≤ 4
              x
≥0, y≥0            
             
a)  First rewrite the inequalities in slope-intercept form:
                  y≤(-3/2)x +3
                  y≤x/2+2 
         
            b)  Enter the right side of those equations opposite Y1 and Y2 respectively and enter 0 opposite Y3.
           c)  Set the WINDOW at Xmin = -1, Ymin=-1, Ymax= the largest value of "b" plus a few units, say 5 in
                 this problem.
                Xmax is a bit more difficult to anticipate.  If there is an equality with a negative slope, I usually
                 make Xmax = 4/3*b/m, round it off to the next largest whole number and add a few units. You can
                  enter the arithmetic opposite Xmax. You might want to press GRAPH and see if all of the
                  corner points of the bounded region are on the screen.
           Now we will enter the inequality signs.
        
a)  Move the cursor to the sign (either equal or inequality) after Y1.  If the inequality symbols do not
                   appear at the bottom of the screen, you will need to start the Inequality App.  Do that by pressing
                   APPS, move the cursor down to Inequal,  or Inequalz for the international version, and press
                   ENTER, ENTER.  The Y= editor screen should be displayed. 
         b)  Place the cursor on the equal sign opposite Y1 and press ALPHA, F3 (ZOOM).  The equal sign
              should have been replaced by the inequality ≤.
             c)  Do the same for Y2; then opposite Y3, press ALPHA, F5 (GRAPH).  The symbol ≥ should have
                   replaced the equal sign before the 0. 
             d)  Now, move the cursor up to the "X" in the upper left corner and press ENTER.  With the cursor on the
                   equal sign opposite X1, press ALPHA, F5 (GRAPH) to enter
;  then enter a zero after that symbol.
             e)  Press GRAPH  to draw and shade the graphs.  
             f)  Press ALPHA, F1, 1 to define the feasible region.
             If you only want to graph, you may stop here.
             At this time we will find the x- and y-values of the corner points.  We will use a method to have the
             calculator determine the corresponding values of the objective function.  If you prefer to determine
             those values manually, just record them without pressing STO so that you can substitute them into
             the objective function.
             a)  Press ALPHA, F3 (ZOOM) and  x=.5, y=2.25 will  be displayed.  Record these values if you choose to evaluate the objective
                   function by hand.  Otherwise, press STO, ENTER.
              b) Press the down arrow to move to the point x=2, y=0 and press STO, ENTER.  Note that different sequences of the cursor
                   keys may be needed to read the points.  It takes a little experimenting.
              c)  Press the right arrow; then then down arrow to read the last point, x=0, y=2. 
             Having the calculator evaluate the objective function.
             
a)  Press STAT, ENTER, to bring up the lists with the stored data.
             
b)  First we will name the list after list INEQY.  We will name it OBJ.  Move the cursor to the name
                    block at the top of that list.  Press 2ND, ALPHA, O, B, J, ENTER.
              Now, we will define the list OBJ.
              c)  With the curosor on the list name, press ALPHA, " (the + key), 2, x(multiply), 2ND, LIST and scroll down to
                    INEQX and press ENTER.
             d)  Press +, 5, x (multiply), 2ND, LIST, scroll to INEQY and press ENTER.  Press ALPHA, ".  You should now have
                   "2*└INEQX+5*└INEQY" at the bottom left of the screen after "OBJ=".  Press ENTER and the values
                   of the objective function for each set of coordinates will be displayed in the OBJ list. 
               Admittedly, this is a rather formidable list of steps, but after a little practice, it becomes quite easy.

      2.  Application Using the Graphical Method:
            Suppose a company manufactures two types of products Good and Better and it has two machines 1 and 2.  The
            company will realize a profit of $5 on each Good item  and $4 on each Better item.  To manufacture a Good
            item, 6 minutes are required on machine 1 and 5  minutes on machine 2. To manufacture a Better item,
            9 minutes are required on machine 1 and 4 minutes on machine 2.  If 5 hours are available on machine 1 and
            3 hours are available on machine 2, how many of each should be manufactured to make the most profit?
            To facilitate graphing, let x = the number of good items and y the number of best items.  Then we want to
            maximize the following profit function:
            P=5x+4y
            We can make a table for our constraints as follows:
            

Machine  G B Totals
1 6 9 300
2 5 4 180

           So, we can write the following constraints:
           6x+9y≤300
           5x+4y≤180
           x ≥ 0, y ≥0
           a)  First rewrite the inequalities in slope-intercept form:
                y≤-2/3x +300/9
                y≤-5/4x+45 
           b)  Press Y= and enter the expressions in Y1 and Y2.  If you wish, you can do the shading by moving the
                cursor over to the left and pressing ENTER until you get the third-quadrant triangle for <, but I prefer to
                just graph the equations.
          c)  Press WINDOW and set the following: Xmin=0, Xmax=50, Ymin=0, Ymax=45.  Press GRAPH and the
               graphs will be displayed. 
          d)  To find the intersection point press 2ND, CALC, 5(Intersect),  ENTER, ENTER.   Move the cursor to the
               approximate intersection and press ENTER again.  The values X=20, Y=20 will be displayed.
          Now, we want to find the point where the feasible region intersects the x- and y-axes.
           e)  Press 2ND, CALC, ENTER (for Value). The values x=0, y= 33.33.... will be displayed.
           f)  Press 2ND, CALC, 2 (for zero) and make sure the equation -5/4x +45 is displayed at the top of the screen. 
                If it is not, press the up arrow of the cursor control.
           f)  Move the cursor down until you are at about x=3 or so and press ENTER.  Move the cursor to some small
                 negative number and press ENTER again. Finally, move the cursor to about zero and press ENTER again.
                 The values X=36, Y=0 will be displayed
          f)  We now have the points (0, 33.3 ), (20, 20), and (36,0) that we want to plug into the objective function. You
              could do this manually fairly easily, but there may be occasions when it is not done so easily by hand. 
              Do the calculations with the calculator as follows:
              1)  Press 0, STO, X (use the [x,T,0,n] button for X), ALPHA, : (the decimal button), 33, STO, ALPHA, Y,
                  ALPHA, :,  5, X, +,  4, ALPHA, Y. You should now have 0
→X:33→Y:5X+4Y.
                   2)  Press ENTER and 132 will be diplayed.
                   3)  Now, press 2ND, ENTER and change the stored values of X and Y, so that you have the following:
                         20→X:20→Y:5X+4Y.  Press ENTER and 180 will be displayed.
                    3)  Now, press 2ND, ENTER and change the stored values of X and Y, so that you have the following:
                         36→X:0→Y:5X+4Y.  Press ENTER and 144 will be displayed.
                   So, the point (20, 20) maximizes the objective function. 

     3. Simplex Method:
         
By far the simplest method for solving simplex problems is with a calculator program.  Later,
         I'll give you a location on my Website where you can find programs to download.  However,
         many students may not solve more than a half-dozen simplex problems in their college career, and some
         professors will require doing simpler simplex problems by hand. So I will include a semi-automated method
         to reduce the drudgery of arithmetic and to reduce errors.
            Ex 1: 
Suppose that we want to maximize the objective function with the function and constraints defined
            as follows:
            z=3x1+2x2+x3
            2x1+ x2 +x3
≤150
                2x1+2x2+8x3
200
            2x1+3x2+x3
≤ 320
            x
1≥0, x2≥0, x3
               We introduce slack variables and have the following simplex tableau:
              
z=3x1+2x2+x3
            x1+ x2 +x3
   s1   s2   s3  |    .
                2      1      1      1    0      0   |150
                2      2      8      0    1      0   |
200
            2     3    1     0  0     1   |
  320
               -3     -2     -1     0   0       0    |0
              Now, we want to solve this simplex problem.  Using the Matrix operations, do that as follows:
              a) Press 2ND, MATRIX, move the cursor to EDIT and press ENTER.
              b) Change the dimensions to 4 x 7 and enter the values into the matrix.  Just enter the number and
                   press ENTER to do that.
               c)  After you have completed entering the information, press 2ND, QUIT to go the home screen.
               d)  Just as a precaution, let's store this in matrix [B] just so we won't have to re-enter everything
                     in case we make a mistake.  So, press 2ND, MATRIX, ENTER, 2ND, MATRIX, 2(B).  You should now
                     have [A]→[B] displayed on the screen.  Press  ENTER and the storing will take place and the matrix
                    will be displayed.
                e) Now, we want to decide where to pivot.  Since -3 is the most negative number, select that column.
                     Now, divide each number in that column into the numbers at the end.  The first number gives
                     150/2=75, which is smaller that the other quotients.  So, we will pivot on that point. 
                 Row Operations:
                      Now, we want to use matrix row operations to leave a 1 in the pivot location and zeros in all other
                      positions in that column.  The syntax of the Operation that we will use is as follows:
                      cc→M:c→A:c→R:*row+(M, [A], A,R)→[A],  where c is a character (number, minus sign, etc.) , M is the
                      row multiplier, A is the number of the row being multiplied, and R is the row being replaced. 
                      Remember that the arrow indicates pressing the STO key, the colons are obtained by pressing
                      ALPHA and the decimal key, and *row+( is obtained by pressing 2ND, MATRIX, moving the
                      cursor to MATH, scroll down to *row+( and pressing ENTER.  The matrix symbol [A], must be
                      obtained by pressing 2ND, MATRIX, ENTER.
                 f)  Now, let's do row 2.  We will have the following on the home screen:
                     -1→M:1→A:2→R:*row+(M, [A], A,R)→[A].  Press ENTER and a zero will appear in the appropriate
                      cell with the other cells in row two modified accordingly.
                 g)  We want a zero in place of the 2 in row 3.  Press 2ND, ENTER and edit the number for R, so that the display
                       now reads as follows:   -1→M:1→A:3→R:*row+(M, [A], A,R)→[A].  This will set up the matrix operation
                       as follows: *row+(-1, [A], 1, 3)→[A].  Press ENTER and the changed matrix will be displayed.
                h)  We want a zero in place of the -3 in row 4.  Press 2ND, ENTER and edit the numbers for M and R,
                      so that you have this:  3/2→M:1→A:4→R:*row+(M, [A], A,R)→[A].  This will set up the matrix operation
                       as follows: *row+(3/2, [A], 1, 4)→[A].   Press ENTER and the changed matrix will be displayed.
                 i)  We want to reduce the first number, 2 to 1.  Normally, we would use the multiply row function to
                      do that, but to reduce key strokes, we're going to do it another way.  So, press 2ND, ENTER and
                      edit the display so that it reads as follows: -1/2→M:1→A:1→R:*row+(M, [A], A,R)→[A].  Press
                      ENTER and the following will be displayed.
                     [[  1     .5      .5      .5    0    0    75]
                      [ 0       1      7       -1    1    0   
50]
                 [0     2     0     -1  0     1  17]

                      [ 0     -.5     .5     1.5   0      0  225]]
                 Now, the pivot point will be on the 1 in column 2.
                 j)  We want a zero in place of the 2 in row 3 Column2..  Press 2ND, ENTER and edit the numbers for M,
                      and R, so that you have this:  -2→M:2→A:3→R:*row+(M, [A], A,R)→[A].  Press ENTER and the changed
                       matrix will be displayed
              
 k)  We want a zero in place of the .5 in row 1, column2.  Press 2ND, ENTER and edit the numbers for M and R,
                      so that you have this:  -.5→M:2→A:1→R:*row+(M, [A], A,R)→[A].  Press ENTER and the changed
                       matrix will be displayed.
                l)  We want a zero in place of the -.5 in row 4, column 2.  Press 2ND, ENTER and edit the numbers for M and R,
                      so that you have this:  .5→M:1→A:4→R:*row+(M, [A], A,R)→[A].  Press ENTER and the changed
                       matrix will be displayed.
                 Since there are no more negative numbers in the last row, the solution is now complete.  The solution
                 you should have is as follows:
                 [[  1        0      -3        1    -.5    0    50]
                    [ 0       1      7        -1      1    0   
50]
               [
0        0   -14     1    -2   1    70]
                    [ 0       0      4        1       0      0  250]]

          3.  Simplex Method with a Calculator Program:
               
There are a number of program available on the Internet.  You can either use one of my programs or do a Google
                 search to find another source.   Here's the URL to go directly to my programs. 
                 http://www.angelfire.com/pro/fkizer/ti_programs/tiprograms.htm
                 Notice that there are different program depending on the type simplex problems you want to solve.  My programs
                 are in text for manual entry and in .8XP format for download and entry by computer.  The instructions are included
                 with the programs.

IV.  Calculus Applications:
       1.  Cost, Average Cost, and Marginal Cost:
           Suppose that a company estimates that the cost (in dollars) of producing x units of a certain product is given by
           C(x) = 200+0.05x+0.0001x².
           Cost of producing 500 units.
           a)  Press Y= and enter the equation 200+.05x +.0002x². Use the [X,T,0,n] key to enter x and the x² key to
                enter the exponent after entering x.
                Press WINDOW and set Xmin at 450; then press ZOOM, 0 (ZoomFit) display the graph. 
           b)  Press 2ND, CALC, ENTER.
           c)  From the displayed graph screen enter 500 opposite s and press ENTER.  The cost of 250 will be displayed.
           Average cost of producing 500 units.
           a)  Press Y= and change the entry so tha tit is as follows: (200+.05x +.0002x²)/x
                Press WINDOW and set Xmin=450; then press ZOOM, 0(ZoomFit) to display the graph.
           b)  Press 2ND, CALC, ENTER and enter 500 opposite x.
           c)  Press ENTER and x=500, y=.5 will be displayed.  The average cost is $0.50.
           Marginal Cost for 500 Units:
            a)  Change the equation back to the cost equation: 200 +.05x+ .0001x².
            b)  From the graphing screen press, 2ND, CALC, 6(dy/dx).
            c)  Enter 500 and press ENTER.  The equation dy/dx =.15 will be displayed.

         2. Average-Cost Minimization:
            
Find the number of units that will minimize the average cost and the corresponding cost.
             We want to find the point where C(x) = C'(x), so we'll need the first derivative: c'(x)=.05+.0002x.
             a)  Enter (200+.005x+.0001x²)/x opposite Y1 and .05+.0002x opposite Y2.
             b)  Set x-min = 450 and x-max = 200.  Press ZOON, 0(ZoomFit) to display the graphs.
               Now find the intersection point.
             c)  Press 2ND, Calc, 5 (intersect).
             d)  From the graph screen, press ENTER, ENTER; then move the cursor approximately to the intersection
                  and press ENTER again.
             e)  The numbers x=1414.21... and y=.332... will be displayed.  So, the number to minimize average cost
                  is approximately 1414 and the cost is approximately .332 dollars.

           3.  Cost Minimization;
                 The cost C(x) of manufacturing x units of a produce is approximated by the following: C(x) =100+10/x+x²/100.
                 Find the number of units that should be manufactured to minimize cost.
                 a)  Press Y= and enter the equation 00+ 10/x + x²/100. Use the [X,T,0,n] key to enter x and the x² key to
                      enter the exponent after entering x.  Press WINDOW and set Xmin =0, Xmax=50, Ymin=90 and Ymax =140.
                 b)  Press 2ND, CALC, 3 (minimum), and move the cursor to the left of the minimum (about 5) and press ENTER.
                      Move the cursor to the right of the minimum (about 20) and press ENTER.  Move the cursor back to the left
                      to approximately the minimum and press ENTER.  The value 10 will be displayed for the number to minimize
                      cost.
           4. Complementary or Substitute Product Determination: 
               Suppose that the demands of two products x sells for $2.50 and product y sells for $3.00 with unit prices p and q
               are given by the following  equations:
               x=8,000 - 0.09p² + 0.08p²
               y=15000 + 0.04p² - 0.3q²
               Are the products substitute complementary or neither.
               The decision-making conditions are summarized in the following table:
               

Partial Derivatives Products x and y
∂f/∂q  > 0 and  ∂f/∂p > 0 Substitute
∂f/∂q < 0 and ∂f/∂p < 0 Complementary
∂f/∂q > 0 and ∂f/∂p < 0 Neither
∂f/∂q < 0 and ∂f/∂p > 0 Neither

              a) Press Y= and enter the following equations:
                  8000 - 0.09*2.50² +0.08x² opposite Y1 and
                  15000+ 0.04*x² -0.3*3².
              b)  Move the cursor to the equal sign opposite Y2 and press ENTER to disable that equation; then press 0 for
                   ZoomFit to GRAPH the equation.
              c)  Press 2nd, CALC, 6.
              d)  ENTER 3 and press ENTER to get 0.48.
               b)  Move the cursor to the equal sign opposite Y1and press ENTER to disable that equation; then move the cursor to Y2
                    and press ENTER to enable that equation.  Press ZoomFit to graph Y2.  
              e)  Press 2nd, CALC, 6.
              d)  ENTER 2.5 and press ENTER to get 0.2
.
                   f)  Since we know that with substitute goods, all other things remaining constant, if the price of increases the demand for the other
                        increases.  In mathematical terms  we can express that as
∂f/∂q  > 0 and  ∂f/∂p > 0.  So, products x and y are substitute.

            5.  Equality of Marginal Product of Labor and Capital.
                 The productivity of a manufacturing organization is given by f(x,y) = 15x0.4 y0.6 .  If the company is utilizing 400 units of
                  labor, what units of capital will make the marginal productivity of labor and the marginal productivity of capital equal? What is
                  the marginal rate at that level?
                  Take the partial derivatieves:  ∂f/∂x=6x-0.6 y0.6 and ∂f/∂y=9x0.4 y--0.4
                  a)  Press Y= and enter 6*400x^-0.6)x ^0.04 opposite Y1.  Opposite Y2, enter 9(400x^0.4)x ^-0.4.
                  b)  Press WINDOW and set Xmin =0, Xmax =700, Ymin =0 and Ymax=10.  Press ENTER to display the graph.
                  c)  Press 2ND, CALC, 5 (Intersect).
                  d)  When the graph appears, press ENTER, ENTER; move the cursor near the intersection and press ENTER again.
                         X=600 and Y=7.652..,  So, 600 units of capital and 400 units of labor will give marginal rates of 7.652.  

            6. Consumers' Surplus:
                 Assume the price-demand equation as follows of p=20-0.05x.  Find the consumers' surplus at $8.
                 Note that we could integrate the function CS=(20-0.05x-8)dx, but I have chosen to simulate an actual integration and evaluate it
                 at both limits.  First, we need to find the upper limit of integration, that is, where the line y=8 intersects the demand curve.
                 a) Press Y= and enter the expression 20-0.05x opposite Y1.  Enter 8 opposite Y2.
                 b) Press WINDOW and set Xmin = 0, Xmax = 250, Ymin=6, Ymax=20.  Press GRAPH to graph the equations.
                 c) Press 2nd, CALC, 5 for Intersect.
                 d) On the graph that appears, press ENTER, ENTER; then move the cursor approximately to the intersection and press ENTER.
                      The intersection, 240 will be the upper limit of integration.
                 Now we want to integrate the areas separately.
                 a) Press Y= and move the cursor to the equal sign opposite Y2 and press ENTER to disable that equation. Press GRAPH.
                 b) Press 2nd, CALC, 7 for integration.  On the graph screen enter 0 for the lower limit and press ENTER.  Then ENTER 240 for
                      the upper limit and press ENTER. The value of the area, 3360, will be displayed.
                 c)  Press Y=, move the cursor to the equal sign opposite Y1 and press ENTER to disable that equation. Press GRAPH.
                 d) Press 2nd, CALC, 7 for integration.  On the graph screen enter 0 for the lower limit and press ENTER.  Then ENTER 240 for
                      the upper limit and press ENTER. The value of the area, 1920, will be displayed.
                 e)  The consumers' surplus, CS=3360-1920 =1440

            7.  Producers' Surplus:
                
Assume the price-supply equation as follows of p=2+0.0002x².  Find the producers' surplus at $20.
                 Note that we could integrate the function CS=(20-2 -0.0002x²)dx, but I have chosen to simulate an actual integration and evaluate it
                 at both limits.  First, we need to find the upper limit of integration, that is, where the line y=20 intersects the supply curve.
                 a) Press Y= and enter the expression 2+0.0002x² opposite Y1.  Enter 20 opposite Y2.
                 b) Press WINDOW and set Xmin = 0, Xmax = 300, Ymin=0, Ymax=25.  Press GRAPH to graph the equations.
                 c) Press 2nd, CALC, 5 for Intersect.
                 d) On the graph that appears, press ENTER, ENTER; then move the cursor approximately to the intersection and press ENTER.
                      The intersection, 268.33, will be the upper limit of integration.
                 Now we want to integrate the areas separately.
                 a) Press Y= and move the cursor to the equal sign opposite Y1 and press ENTER to disable that equation. Press GRAPH.
                 b) Press 2nd, CALC, 7 for integration.  On the graph screen enter 0 for the lower limit and press ENTER.  Then ENTER 268.33 for
                      the upper limit and press ENTER. The value of the area, 2146.66, will be displayed.
                 c)  Press Y=, move the cursor to the equal sign opposite Y1 and press ENTER to disable that equation. Press GRAPH.
                 d) Press 2nd, CALC, 7 for integration.  On the graph screen enter 0 for the lower limit and press ENTER.  Then ENTER 268.33 for
                      the upper limit and press ENTER. The value of the area, 5366.6, will be displayed.
                 e)  The suppliers' surplus, CS=5366.6-2146.66 =3219.94.

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