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1 Mathematics for Management Science Notes 08 prepared by Professor Jenny Baglivo Graphical representations As an introduction to the calculus of two-variable functions (f(x ;x 2 )), consider two graphical representations of these functions: ffl Surfaces in three-space: Collections of points of the form (x ;x 2 ;f(x ;x 2 )). ffl Contour plots in two-space: Collections of curves satisfying f(x ;x 2 ) = C for several different values of the constant C. Example. Consider the two-variable function f(x ;x 2 )=4 x 2 x2 2. x2 2 f x2 0 - x x ffl The plot on the left is the surface when 2» x» 2 and 2» x 2» 2. ffl The plot on the right shows contours (level curves) corresponding to f(x ;x 2 )= 2,, 0,, 2, 3, 4. Example 2. Consider the two-variable function f(x ;x 2 )=x 3 + x3 2 3x x 2. f x 2-2 x2 x x ffl The plot on the left is the surface when» x» 2 and» x 2» 2. ffl The plot on the right shows contours (level curves) corresponding to f(x ;x 2 )=, 0,, 2, 3, 4, 5.

2 Partial derivatives The partial derivative in the x -direction is the function you get by differentiating f(x ;x 2 ) with respect to x for fixed x 2. The notation is as follows: f (x ;x 2 ) ;x 2 : The partial derivative in the x 2 -direction is the function you get by differentiating f(x ;x 2 ) with respect to x 2 for fixed x. The notation is as follows: f 2 (x ;x 2 ) ;x 2 2 : Example, continued: Let f(x ;x 2 )=4 x 2 x2 2. a. Compute f (x ;x 2 )andf 2 (x ;x 2 ). b. Compute f( :5; ), f ( :5; ), and f 2 ( :5; ). x x 2

3 Example 2, continued: Let f(x ;x 2 )=x 3 + x3 2 3x x 2. a. Compute f (x ;x 2 )andf 2 (x ;x 2 ). b. Compute f(:5; ), f (:5; ), and f 2 (:5; ) x x Critical pairs, critical points The pair (x ;x 2 ) is said to be a critical pair if f (x ;x 2 )=0 and f 2 (x ;x 2 )=0: The triple (x ;x 2 ;f(x ;x 2 )) is said to be a critical point if the pair (x ;x 2 ) is a critical pair. 3

4 Examples and 2, continued: ffl Find the critical pairs and critical points of f(x ;x 2 )=4 x 2 x2 2. ffl Find the critical pairs and critical points of f(x ;x 2 )=x 3 + x3 2 3x x 2. 4

5 Second derivative test The second derivative testsfor finding local maxima and local minima of two-variable functions uses:. The second partial derivative inthex -direction: f (x ;x 2 ) (x ;x 2 ) 2 f(x ;x The second partial derivative inthex 2 -direction: f 22 (x ;x 2 ) 2(x ;x 2 ) 2 f(x ;x 2 ) The mixed second partial derivatives: f 2 (x ;x 2 ) (x ;x 2 ;x 2 f 2 (x ;x 2 ) 2(x ;x 2 ;x @x 2 f(x ;x 2 f(x ;x 2 : and For functions in this course, we will get f 2 (x ;x 2 )=f 2 (x ;x 2 ). The second partial derivatives are combined into a discriminant function: D(x ;x 2 ) = f (x ;x 2 ) f 22 (x ;x 2 ) f 2 (x ;x 2 ) f 2 (x ;x 2 ) : Now, suppose that (x ;x 2 ) is a critical pair for the function f, andd(x ;x 2 ) 6= 0. Then,. If D(x ;x 2 ) > 0 and f (x ;x 2 ) > 0andf 22 (x ;x 2 ) > 0, then there is a local minimum at the critical pair (x ;x 2 ). 2. If D(x ;x 2 ) > 0 and f (x ;x 2 ) < 0 and f 22 (x ;x 2 ) < 0, then there is a local maximum at the critical pair (x ;x 2 ). 3. If D(x ;x 2 ) < 0, then there is a saddle point at(x ;x 2 ). There is a saddle point at (x ;x 2 ) if in some directions there is a local minimum at (x ;x 2 ) and in some directions there is a local maximum at (x ;x 2 ). 5

6 Example, continued: Let f(x ;x 2 )=4 x 2 x2 2. ffl Compute the second partials (f, f 2, f 2, f 22 ) and the discriminant function (D). ffl Use the second derivative test to verify that there is a local maximum at (0; 0). 6

7 Example 2, continued: Let f(x ;x 2 )=x 3 + x3 2 3x x 2. ffl Compute the second partials (f, f 2, f 2, f 22 ) and the discriminant function (D). ffl Use the second derivative test to verify that there is a local minimum at (; ) and a saddle point at(0; 0). 7

8 Example 3. Lawn King, Inc., manufactures two types of lawn mower. One is a riding mower and the other is a standard walking power mower. Lawn King is interested in establishing a pricing policy for the two mowers that will maximize the total profit for the product line. Sales for these products are not independent. Over time, Lawn King has observed that an increase in the price of the riding mower is usually accompanied by a decrease in the number of riding mowers sold and an increase in the number of walking mowers sold. Similarly, increases in the price of the walking mower have been accompanied by a decrease in sales of the walking mower and an increase in sales of the riding mower. Economists would say that the two types of lawn mower are substitutable products. After studying the relationships between sales prices and quantities sold, Lawn King has established the following expressions: Demand (in thousands) for walking mowers = 95 2p +0:7p 2 Demand (in thousands) for riding mowers = :3p 0:5p 2 where p is the price of the walking mower and p 2 is the price for the riding mower. Further information, prepared by the accounting department, is available on the total cost of producing each type of mower: Cost (in thousands $) for walking mowers = q Cost (in thousands $) for riding mowers = q 2 where q is the quantity in thousands of walking mowers produced and q 2 is the quantity in thousands of riding mowers produced. As a preview of the solution, the plot on the left shows profit as a function of p and p 2 when profit is non-negative and the plot on the right shows contours for f(p ;p 2 ) equal to 0, 2000, 4000, 6000, 8000, and 9000 (in thousands of dollars). p f p2 300 p p 8

9 a. Express profit as a function of p and p 2 : f(p ;p 2 ). b. Find the critical pairs of f(p ;p 2 ). 9

10 c. Use calculus to determine the values of p and p 2 to maximize profit. d. Find the production levels (q and q 2 ) to maximize profit and find the maximum profit. 0

11 The form of a quadratic polynomial function in two variables is as follows: f(x; y) = c x 2 + c 2 y 2 + c 3 xy + c 4 x + c 5 y + c 6 where the c i 's are constants, and either c 6=0orc 2 6=0orc 3 6=0. Quadratic polynomial functions have a unique critical pair. If there is a local min at the critical pair, then there is also a global min at the pair. If there is a local max at the critical pair, then there is also a global max at the pair. Constrained optimization The goal in this unit is to give anintroduction to sensitivity analysis for nonlinear programming problems in the following simple setting: Optimize f(x ;x 2 ) subject to ax + bx 2 = C where C is a constant. The linear equality corresponds to a bound constraint. The method we will use is similar to the one we used in the linear case, simplifies the approach used in the book, and computes an answer with the correct sign. Example 4 Green Lawns, Inc. provides a lawn fertilizer and weed control service. The company provides four treatments of fertilizer and weed control chemical to its subscribers each year. Green Lawns is adding a special aeration treatment as a low-cost extra service option, which it hopes will help attract new customers. Management is planning to promote this new service in two media: radio and direct-mail advertising. A budget of $2000 is to be used on this promotional campaign over the next quarter. Based on past experience in promoting its other services, Green Lawns has been able to obtain an estimate of the relationship between sales and the amount spent on promotion in these two media: S(x ;x 2 ) = 2x 2 0x2 2 8x x 2 +8x +34x 2 where x represents thousands of dollars spent on radio advertising, x 2 represents thousands of dollars spent on direct-mail advertising, and S(x ;x 2 ) is the total sales in thousands of dollars. Green Lawns would like to determine a promotional strategy that will lead to maximum sales subject to the restriction provided by the promotional budget. To compute the solution by hand, () Observe that x + x 2 =2(thousand dollars) or x 2 =2 x. Thus, total sales can be written as a function of x only: f(x )=S(x ; 2 x ) = 2x 2 0(2 x ) 2 8x (2 x )+8x + 34(2 x ) = 4x 2 + 8x +28 :

12 (2) f(x ) has a unique critical value at x = since f 0 (x )= 8x +8=0 =) x =: Further, the critical value corresponds to a local maximum since f 00 () = 8 < 0. (3) Since f(x ) is a quadratic polynomial, the local maximum is also a global maximum. Solution: Green Lawns, Inc. should spend one thousand dollars each on radio and directmail advertising. The maximum profit is 32 thousand dollars. Sensitivity analysis in the Green Lawns example Consider sensitivity toa change in the total promotional budget. It is common practice to let represent the additional funds available (in thousands of dollars). Then the new constraint becomes x + x 2 = 2 + : Note that at the current solution x =, =0,andx 2 =2+ x =. Step. Construct a function of by substituting x = and x 2 =2+ x =2+ =+ into the total sales function: f( ) =S(; + ) = 2 0( + ) 2 8( + ) ( + ) = : Step 2. Compute the rate of change of sales with respect to. That is, compute f 0 ( ): f 0 ( ) = Step 3. Then dsales/drhs at the optimal solution is f 0 (0): f 0 (0) = If the promotional budget is increased by one thousand dollars, then sales will increase by approximately f 0 (0) = thousand dollars. 2

13 Footnotes on sensitivity analysis:. The letter is for Lagrange, the mathematician who formalized the method. 2. The value f 0 (0) is often referred to as the Lagrange multiplier. 3. In general, we will obtain a non-zero Lagrange multiplier for each bound constraint in a non-linear programming problem. (Constraints are usually inequalities, and not equalities.) A spreadsheet solution, sensitivity report, and formulas for the Lawn Kings example are given at the end of the last set of notes (notes07). Example 5 Consider the following non-linear programming problem: Model: Minimize COST(x ;x 2 ) = 2x 2 20x +2x x 2 + x 2 2 4x Subject to: x + x 2 = 0 x ; x 2 0 a. Use calculus to find the optimal solution. 3

14 b. Find dcost/drhs, where RHS represents the righthand side of the equality constraint. 4

15 Example 6 The directors of the TMC Corporation are trying to determine how to allocate their research staff of thirty-six engineers for the coming year. Three different projects are under consideration. The directors believe that the success of each project depends in part on the number of engineers assigned, with probabilities shown in the following graph: Probability x Specifically, if x engineers are assigned to: ffl Project (wide dashes), then the probability of success is ffl Project 2 (small dashes), then the probability of success is ffl Project 3 (solid), then the probability of success is x (x+4). x (x+2). x (x+9). The directors estimate that, if successful, the net present values of the three projects are $800,000, $700,000, $,200,000, respectively. TMC would like to determine the resource allocation that will maximize the total expected net present value: Expected NPV (thous) = 800 X (X +2) +700 X 2 (X 2 +9) +200 X 3 (X 3 +4) where X i is the number of engineers assigned to project i, i =; 2; 3. Decision Variables: X i is the number of engineers assigned to project i, fori =,2,3. Model: Maximize: Expected NPV (thous) = 800 (X +2) +700 X2 (X 2+9) +200 X3 Subject to: X + X 2 + X 3» 36 X ; X 2 ; X 3 0 X (X 3+4) 5

16 a. If all 36 engineers are assigned to project, then the expected NPV is. b. If all 36 engineers are assigned to project 2, then the expected NPV is. c. If all 36 engineers are assigned to project 3, then the expected NPV is. Refer to the spreadsheet solution at the end of the last set of notes (notes07). d. Clearly state the optimal solution. e. Interpret the Lagrange multiplier for the first constraint. 6