Question

A company manufactures two types of electric hedge trimmers, one of which is cordless. The cord-type trimmer requires 2 hours to make, and the cordless model requires 4 hours. The company has only 800 work hours to use in manufacturing each day, and the packaging department can package only 300 trimmers per day. If the company profits for the cord-type model for $28.50 and the cordless model for $57.00, how many of each type should it produce per day to maximize profits? Scenario #1 (with the smaller number of cord-type trimmers): corded models cordless models Scenario #2 (with the larger number of cord-type trimmers): corded models cordless models

Answer 1

Let x cordless and y cord-type trimmers be manufactured daily. Since the cord-type trimmer requires 2 hours to make, and the cordless model requires 4 hours , and since the company has only 800 work hours to use in manufacturing each day, hence 2y+4x ? 800 or, y+2x ? 400 or, y ? -2x +400…(1)

Also, since the packaging department can package only 300 trimmers per day, therefore, x +y ? 300 or, y ? -x+300…(2).

The company makes a profit of $28.50 on the cord-type model and $57.00 on the cordless model so that the company’s daily profit is P(x) = 28.50y +57x…(3).

A graph of the lines L₁: y = -2x +400 (in red) and L₂: y = -x+300 (in blue)is attached. The feasible region is on or below the lines L₁ and L₂ in the 1^st quadrant ( as both x,y must be ? 0).These 2 lines intersect at the point where x = 100 and y = 200. At this point, both the conditions (1) and (2) are satisfied.

Thus, P(x) will be maximum when x = 100 and y = 200. Then P(x) = 28.50*200+57*100 = 5700+5700 = $11400.

Thus, 100 cordless and 200 cord type trimmers should be manufactured daily to maximize profits.