a firm is specialized in building and selling family homes. the firm offers two

Question

a firm is specialized in building and selling family homes. the firm offers two basic models - A and B. Model A requires 4000 labor hours, 2 tons of stone and 2000 board feet of lumber. Model B houses require 10000 labor hours, 3 tons of stone and 2000 board feet lumber. due to long lead times of supplies and manpower in the areafirm is forced to rely on existing supplies for upcoming building season. it has 400,000 hours of Labour, 150 tons of stone and 200,000 board feet lumber. what mix of model A and B houses should the firm construct if model A yields a profit of $3000 per unit and model B yields $6000 per unit. Assume that the firm will be able to sell all the units it builds.
1) formulate objective function and constraints
2) determine optimal quantities of model A and B and compute the resulting profit.
3) what is the optimum point
4) solve the equations for optimum points.

Please answer all the parts for a thumbs up. Please do not attempt if you are not sure of answers. please provide correct answers to all the questions. thank you.

Explanation / Answer

1) Formulating the objective function and constraints:

Maximize Z = 3000A + 6000B

S.T, Labor = 4000A + 10000B <  400,000 labor hours

Stone = 2A + 3B < 150 tons

Lumber = 2000A + 2000B <  200,000 board feet

A,B >  0

3) The optimum point of the intersection is at stone and labor. Stone and Labor constraints intersect at a point.

2) Determine the optimal quantities of models A and B and compute the profit:

Labor 4000A + 10,000B = 400,000

-2000 * (Stone 2A + 3B) = 150

= 4000B = 100,000

Therefore, B = 25

Substitute B = 25 in one of the equations and solve for A

2A + 3(25) = 150

Therefore A = 37.5

Substitute A and B in Z equation

Z = 3000 (37.5) + 6000 (25) = 262,500

4) Solve equations for optimal points :

The optimal points are -

A = 0, B = 40 (found by inspection)

A = 37.5, B = 25 (found by using simultaneous equations)

A = 80, B = 0 (found by inspection)

From first set of points, A= 0, B = 40

Z = 3000 (0) + 6000 (40) = 240,000

From second set of points A = 37.5, B = 25

Z = 3000(37.5) + 6000(25) = 262,500

From third set of points A = 80, B = 0

Z = 3000(80) + 6000(0) = 240,000

The best value of Z = 262,500. Therefore, the optimal points are A = 37.5 and B = 25

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