8-9 - Exercise 8-9 Revenue function Cost function Profit...

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Exercise 8-9 Revenue function 40x + 42y Cost function 30x + 30y Profit function 10x + 12y Max profit Max w(x,y) = 10x + 12y s.t. x + y ≤ 400 Machine hours x + 2y ≤ 500 Capacity x, y ≥ 0 Part A To find max with both x and y, find point where constraints intersect Rearrange: x + y = 400 x + y – 400 = 0 x + 2y = 500 x + 2y – 500 = 0 Set equal to each other: x + 2y – 500 = x + y – 400 y – 500 = -400 y = 100 therefore x = 300 Both x and y All y All x x = 300 x = 0 x = 400 y = 100 y = 250 y = 0 w(x,y) = 4200 w(x,y) = 3000 w(x,y) = 4000 Both constraints binding Constraint 2 binding Constraint 1 binding So optimal production is x = 300 and y =100 PART B Remember that shadow prices on constraints are > 0 because the constraints are binding There are two different methods to get shadow prices in linear programs. We’ll use one method for the first constraint and another method for the second constraint. Shadow Price on Constraint 1 Remember our model: Max w(x,y) = 10x + 12y s.t. x + y ≤ 401 Machine hours x + 2y ≤ 500 Capacity x, y ≥ 0 If we have one more machine hour, we can increase production. We can’t increase
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