Q. 27 - Precalculus Enhanced with Graphing Utilities

Question

Maximizing Profit on Ice Skates A factory manufactures two kinds of ice skates: racing skates and figure skates. The racing skates require 6 work-hours in the fabrication department, whereas the figure skates require 4 work-hours there. The racing skates require 1 work-hour in the finishing department, whereas the figure skates require 2 work-hours there. The fabricating department has available at most 120 work-hours per day, and the finishing department has no more than 40 work-hours per day available. If the profit on each racing skate is $\backslash (10$ and the profit on each figure skate is $\backslash )12$ , how many of each should be manufactured each day to maximize profit? (Assume that all skates made are sold.)

Step-by-Step Solution

Answer

The maximum profit is obtained when 10 racing skates and 15 figure skates are manufactured each day.

Step 1. Given information

Let the number of racing skates be x and the number of figure skates be y.
If P denotes the total profit obtained from both these skates, then P can be expressed as $P=10x+12y$ This expression is called the objective function.
We have to maximize the value of P subject to the two constraints x and y. 

Given the number of work hours required for fabricating racing skates and figure skates, we get the first constraint as $6x+4y\le 120$

Also, given are the number of work hours required for finishing racing skates and figure skates. We get the other constraint as  $x+2y\le 40$

Since the number of figure skates and racing skates cannot be a negative number, we have two more constraints, $x\ge 0$ and $y\ge 0$.

Step 2. Graph the inequalities we obtained and label the corner points.

The shaded portion of the graph represents the set of feasible points.

Step 3. Find the value of the objective function at each corner point.

Vertex	Value of profit $P=\$10x+\$12y$
$(0,20)$	$P=\$10\left(0\right)+\$12\left(20\right)=\$240$
$(10,15)$	$P=\$10\left(10\right)+\$12\left(15\right)=\$280$
$(20,0)$	$P=\$10\left(20\right)+\$12\left(0\right)=\$200$

From the table, we can see that profit is maximum at $\$280$ for $x=\$10$ and $y=\$12$.
Therefore, the maximum profit is obtained when 10 racing skates and 15 figure skates are manufactured each day.