Question

A company manufactures x units of Product A and y units of Product B, on two machines, I and II. It has been determined that the company will realize a profit of $2/unit of Product A and a profit of $4/unit of Product B. To manufacture a unit of Product A requires 6 min on Machine I and 5 min on Machine II. To manufacture a unit of Product B requires 9 min on Machine I and 4 min on Machine II. There are 5 hr of machine time available on Machine I and 3 hr of machine time available on Machine II in each work shift. How many units of each product should be produced in each shift to maximize the company's profit?

(x, y)

=

What is the optimal profit? (Round your answer to the nearest whole number.)
$

Answer 1

Let x units of Product A and y units of Product B be manufactured in each shift for optimal profit.

Manufacturing of x units of Product A will require 6x min on Machine I and 5x min on Machine II.

Manufacturing of y units of Product B will require 9y min on Machine I and 4y min on Machine II.

Since 5 hr of machine time is on Machine I in each work shift, hence 6x+9y ≤ 5*60 or, 6x+9y ≤ 300 or, 2x+3y ≤ 100…(1).

Also, since 3 hr of machine time available on Machine II in each work shift, hence 5x+4y ≤ 3*60 or, 5x+4y ≤ 180…(2)

The company’s profit per shift is P(x,y) = 2x+4y.

We have to maximize P(x,y) subject to the constraints (1) and (2).

A graph of the lines 2x+3y = 100 (in red) and 5x+4y= 180 ( in blue) is attached. The feasible region is on and below both the lines in the 1^st quadrant ( as both xz and y cannot be negative).

The two lines intersect at the point (20,20). Here, both the constraints are satisfied. Thus, 20 units of each product should be produced in each shift to maximize the company's profit.

The company’s optimal profit is $ (2*20+4*20) = $ 120 per shift.