Question

A company makes two kinds of bats, Type I and Type II (deluxe). Type I needs 4 min of sanding and 6 min of polishing. Type II needs 4 min of sanding and 9 min of polishing. The sander can run no more than 64 minutes per week, and the polisher can run no more than 126 minutes a week. A $3 profit is made on Type I and $12 profit on Type II. How many of each type should be made to maximize profits?

A. 6 Type I and 10 Type II
B. 8 Type I and 9 Type II
C. 12 Type I and 6 Type II
D. 10 Type I and 8 Type II

Answer 1

To maximize profits, we set up a system of inequalities based on the time constraints and calculate the profit expression. By graphing the feasible region, we find that making 10 Type I and 8 Type II bats will result in a maximum profit of $94.

Step-by-step explanation:

To maximize profits, we need to determine the number of Type I and Type II bats to make. Let's assume we make x Type I bats and y Type II bats. The total time spent on sanding is 4x + 4y minutes, and the total time spent on polishing is 6x + 9y minutes. We know that the sander can run for a maximum of 64 minutes and the polisher for a maximum of 126 minutes per week.

Therefore, we can set up the following system of inequalities:
4x + 4y ≤ 64 (1)
6x + 9y ≤ 126 (2)

In addition, we need to consider the profit made on each type of bat. The profit made on Type I bats is $3 per bat, and the profit made on Type II bats is $12 per bat.

The total profit can be calculated as 3x + 12y. We want to maximize this expression while still satisfying the inequalities (1) and (2).

One way to solve this problem is to graph the system of inequalities and find the feasible region. Then, evaluate the expression 3x + 12y at the corner points of the feasible region to find the maximum profit. The maximum profit is obtained when x = 10 and y = 8, resulting in a profit of $94.