A company makes square and triangular boxes. Square boxes take 2 minutes and yield a profit of K4 each. Triangular boxes take 3 minutes and yield a profit of K5 each. The client wants at least 25 boxes in total, with at least 5 of each type. The company has one hour (60 minutes) to make the boxes. The goal is to determine the number of each type of box to maximize profit.

1. Problem Description

A company makes square and triangular boxes. Square boxes take 2 minutes and yield a profit of K4 each. Triangular boxes take 3 minutes and yield a profit of K5 each. The client wants at least 25 boxes in total, with at least 5 of each type. The company has one hour (60 minutes) to make the boxes. The goal is to determine the number of each type of box to maximize profit.

2. Solution Steps

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

Objective function:

Maximize profit $P = 4x + 5y$

Constraints:

Total boxes: $x + y \ge 25$

Minimum square boxes: $x \ge 5$

Minimum triangular boxes: $y \ge 5$

Time constraint: $2x + 3y \le 60$

$x, y$ are non-negative integers.

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

First, analyze the feasible region. We consider integer values for $x$ and $y$.

From $x + y \ge 25$, we have $y \ge 25 - x$.

From $x \ge 5$ and $y \ge 5$, we have those lower bounds.

From $2x + 3y \le 60$, we have $y \le (60 - 2x) / 3$.

We seek integer solutions.

Possible corner points of the feasible region occur where constraint lines intersect.

Intersection of $x + y = 25$ and $2x + 3y = 60$:

Multiply the first equation by 2 to get $2x + 2y = 50$. Subtract this from $2x + 3y = 60$ to get $y = 10$. Then $x = 25 - 10 = 15$. So, $(15, 10)$ is a possible solution.

Intersection of $x = 5$ and $x + y = 25$:

Then $y = 25 - 5 = 20$. So, $(5, 20)$ is a possible solution.

Intersection of $y = 5$ and $x + y = 25$:

Then $x = 25 - 5 = 20$. So, $(20, 5)$ is a possible solution.

Intersection of $x = 5$ and $2x + 3y = 60$:

$2(5) + 3y = 60$, $10 + 3y = 60$, $3y = 50$, $y = 50/3 \approx 16.67$. Since $y$ must be an integer, we can consider $y=16$ or $y=17$. Since $2(5) + 3(16) = 10 + 48 = 58 < 60$ and $5 + 16 = 21 < 25$ is false, so $(5,16)$ is not feasible.

$2(5) + 3(17) = 10 + 51 = 61 > 60$ so $(5,17)$ is not feasible.

Since we need to satisfy $x+y>=25$, try $5+y=25$, then $y=20$. $2(5) + 3(20) = 70 > 60$ which is also not feasible.

So we look for the smallest $y$ where $x+y>=25$

If $x = 5$, $2x + 3y \le 60$ and $x + y \ge 25$. Then $3y \le 50$ so $y \le 16.67$. We also need $y \ge 20$. No integer solution.

Intersection of $y = 5$ and $2x + 3y = 60$:

$2x + 3(5) = 60$, $2x + 15 = 60$, $2x = 45$, $x = 22.5$. Since $x$ must be an integer, we can consider $x=22$ or $x=23$.

$2(22) + 3(5) = 44 + 15 = 59 < 60$ and $22+5=27 \ge 25$, so (22,5) is feasible.

$2(23) + 3(5) = 46 + 15 = 61 > 60$ which is not feasible.

Try $x=22$, $y=5$.

$4(22) + 5(5) = 88+25 = 113$

Let's check the earlier intersection of the first two lines. $(15, 10)$

$4(15) + 5(10) = 60 + 50 = 110$

$2(15) + 3(10) = 30 + 30 = 60 \le 60$

$15 + 10 = 25 \ge 25$

$15 \ge 5$ and $10 \ge 5$. So, $(15, 10)$ is a possible solution.

If $x = 6$, then $y \ge 19$ and $2(6) + 3y \le 60$, so $3y \le 48$, so $y \le 16$. There are no valid $y$.

The other intersections yielded $(5,20)$. Check constraint. $2(5) + 3(20) = 10 + 60 = 70 > 60$. Not feasible. Try $2x+3y=60$ such that $y\ge 5$.

If $y=5$, $2x+15=60$, $2x=45$. No integer solution.

If $y=6$, $2x+18=60$, $2x=42$, $x=21$. $x+y=27 \ge 25$, $x\ge 5, y \ge 5$,

$P=4x+5y = 4(21)+5(6) = 84+30=114$

Let $x=21$ and $y=6$. This satisfies all conditions:

Total boxes: $21+6 = 27 \ge 25$

Minimum boxes: $21 \ge 5$ and $6 \ge 5$

Time constraint: $2(21)+3(6) = 42+18 = 60 \le 60$

The value of objective is $P= 4(21)+5(6) = 84+30 = 114$

3. Final Answer

The best combination is 21 square boxes and 6 triangular boxes.