A company makes square boxes (2 minutes to make, profit of K4) and triangular boxes (3 minutes to make, profit of K5). The client wants at least 25 boxes total, with at least 5 of each type. The company has one hour (60 minutes) to make the boxes. We need to find the combination of square and triangular boxes that maximizes profit.

1. Problem Description

A company makes square boxes (2 minutes to make, profit of K4) and triangular boxes (3 minutes to make, profit of K5). The client wants at least 25 boxes total, with at least 5 of each type. The company has one hour (60 minutes) to make the boxes. We need to find the combination of square and triangular boxes that maximizes profit.

2. Solution Steps

Let $x$ be the number of square boxes and $y$ be the number of triangular boxes.

The objective function to maximize is the profit:

$P = 4x + 5y$

The constraints are:

$x + y \ge 25$ (at least 25 boxes)

$x \ge 5$ (at least 5 square boxes)

$y \ge 5$ (at least 5 triangular boxes)

$2x + 3y \le 60$ (60 minutes available)

$x$ and $y$ are non-negative integers.

We want to find integer values of $x$ and $y$ that satisfy all constraints and maximize $P$.

First, consider the constraint $2x + 3y \le 60$.

If $x = 5$, then $10 + 3y \le 60$, which means $3y \le 50$, so $y \le \frac{50}{3} \approx 16.67$. Since $y$ must be an integer and $y \ge 5$, we have $5 \le y \le 16$.

We also have $x + y \ge 25$, so $5 + y \ge 25$, which implies $y \ge 20$. But we know $y \le 16$, so $x=5$ is not possible.

Let's consider values of x that give integer solutions.

Case 1: $x + y = 25$. Then $x = 25 - y$. Substitute into $2x + 3y \le 60$:

$2(25 - y) + 3y \le 60$

$50 - 2y + 3y \le 60$

$y \le 10$

Since $y \ge 5$, we have $5 \le y \le 10$. Also, $x \ge 5$, so $25 - y \ge 5$, meaning $y \le 20$.

So $5 \le y \le 10$. We want to maximize $P = 4x + 5y = 4(25-y) + 5y = 100 + y$.

To maximize $P$, we take the largest possible value for $y$, which is

1

0. If $y = 10$, then $x = 25 - 10 = 15$.

Check the constraints: $15 + 10 = 25 \ge 25$, $15 \ge 5$, $10 \ge 5$, $2(15) + 3(10) = 30 + 30 = 60 \le 60$.

The profit is $P = 4(15) + 5(10) = 60 + 50 = 110$.

Case 2: $2x + 3y = 60$. Then $2x = 60 - 3y$, so $x = 30 - \frac{3}{2}y$.

For $x$ to be an integer, $y$ must be even. Since $y \ge 5$, we must have $y \ge 6$.

$x \ge 5$ implies $30 - \frac{3}{2}y \ge 5$, so $25 \ge \frac{3}{2}y$, meaning $y \le \frac{50}{3} \approx 16.67$.

$x + y \ge 25$ implies $30 - \frac{3}{2}y + y \ge 25$, so $5 \ge \frac{1}{2}y$, meaning $y \le 10$.

Combining these, since $y$ must be even, we have $6 \le y \le 10$.

If $y = 6$, then $x = 30 - \frac{3}{2}(6) = 30 - 9 = 21$. $x+y = 27 \ge 25$, $P = 4(21) + 5(6) = 84 + 30 = 114$.

If $y = 8$, then $x = 30 - \frac{3}{2}(8) = 30 - 12 = 18$. $x+y = 26 \ge 25$, $P = 4(18) + 5(8) = 72 + 40 = 112$.

If $y = 10$, then $x = 30 - \frac{3}{2}(10) = 30 - 15 = 15$. $x+y = 25 \ge 25$, $P = 4(15) + 5(10) = 60 + 50 = 110$.

Now we need to find other possible combinations where $2x+3y < 60$ and $x+y > 25$.

If $2x+3y = 59$, then $2x = 59 - 3y$, so this has no integer solutions for $x$ since $59-3y$ must be even.

If $2x+3y = 58$, $x+y = 26$, then $2x+2y = 52$. So $y = 6$, $x = 20$. Then profit $4(20)+5(6) = 80+30=110$.

If $2x+3y = 57$, then $x$ needs to be integer so $57-3y$ must be even, which means $y$ must be odd.

Possible profits:

x=21, y=6, P =

1

1

4. x=20, y=7, $2(20)+3(7) = 40+21 = 61$, impossible.

x=18, y=8, P=

1

1

2.

So the maximum profit is 114, when $x = 21$ and $y = 6$.

3. Final Answer

The best combination is 21 square boxes and 6 triangular boxes, which yields a profit of K

1

1

4.