Question

1) The Outdoor Furniture Corporation manufactures two products, benches and picnic tables, for use in yards and parks. The firm has two main resources: its carpenters (labor force) and a supply of redwood for use in the furniture. During the next production cycle, 1200 hours of labor are available under a union agreement. The firm also has a stock of 3500 board feet of good-quality redwood. Each bench that Outdoor Furniture produces requires 4 labor hours and 10 board feet of redwood; each picnic table takes 6 labor hours and 35 board feet of redwood. Completed benches will yield a profit of $9 each, and tables will result in a profit of $20 each. How many benches and tables should Outdoor Furniture produce to obtain the largest possible profit? Use the graphical LP approach (corner point).

Answer 1

MAX 2 = 9x1 + 20x2 subject to 10x1 + 35x2 <= 3500 4x1 + 6x2 <= 1200 and x1,x2 >= 0 Solution: Problem is MAX Z, = 9X1 + 20X2 subject to 10X1 + 35x25 3500 4 X1 + 6 X2 < 1200 and x1,x220;

Hint to draw constraints 1. To draw constraint 10x1 + 35x2 < 3500 + (1) Treat it as 10xı + 35x2 = 3500 When xı = 0 then x2 = ? = 100) + 35x2 = 3500 = 35x2 = 3500 3500 = x2 = 35 = 100 When x2 = 0 then x1 = ? = 10x7 + 35(0) = 3500 = 10x1 = 3500 = x1 = 10 3500 - 350 x2 0 350 |x2 100 0

2. To draw constraint 4x1 + 6x2 < 1200 - (2) Treat it as 4xı + 6x2 = 1200 Whenx = 0 then x2 = ? = 4(0) + 6x2 = 1200 = 6x2 = 1200 1200 = x2 = 6 = 200 When x2 = 0 then xy = ? = 4x + 6(0) = 1200 = 4x2 = 1200 1200 =*= = 300 X1 O 300 x2 200 o

10:50 F110 RH 9 10 mm mm \

Objective function value z = 9x1 + 20x2 90 + 2000) = 0 The value of the objective function at each of these extreme points is as follows: Extreme Point Coordinates Lines through Extreme Point (X1X2) 3+x20 0(0,0) 4 + x2 > 0 2 - 4x + 6x2 < 1200 A(300,0) 4+x220 1 - 10x1 + 35x2 S 3500 B(262.5, 25) 2 - 4x1 + 6x2 < 1200 1 – 10x7 + 35x7 S 3500 CO, 100) 3 → X20 9(300) + 2000) = 2700 9(262.5) +20(25) = 2862.5 9(0) +20(100) = 2000 The maximum value of the objective function z = 2862.5 occurs at the extreme point (262.5, 25). Hence, the optimal solution to the given LP problem is : x1 = 262.5.x, = 25 and max z = 2862.5.

Objective function value z = 9x1 + 20x2 90 + 2000) = 0 The value of the objective function at each of these extreme points is as follows: Extreme Point Coordinates Lines through Extreme Point (X1X2) 3+x20 0(0,0) 4 + x2 > 0 2 - 4x + 6x2 < 1200 A(300,0) 4+x220 1 - 10x1 + 35x2 S 3500 B(262.5, 25) 2 - 4x1 + 6x2 < 1200 1 – 10x7 + 35x7 S 3500 CO, 100) 3 → X20 9(300) + 2000) = 2700 9(262.5) +20(25) = 2862.5 9(0) +20(100) = 2000 The maximum value of the objective function z = 2862.5 occurs at the extreme point (262.5, 25). Hence, the optimal solution to the given LP problem is : x1 = 262.5.x, = 25 and max z = 2862.5.