Toy Factory Problem

Let $c$ be the number of cars produced, let $m$ be the number of motorcycles produced, and let $b$ be the number of boats produced. The goal is to maximize the objective function:

$f(c,m,b)=20c+15m+25b$

Subject to the constraints:

$\begin{cases}\begin{aligned} 2c+m+2b &\le 8 &\text{Casting} \\ 2c+2m+3b &\le 12 &\text{Assembly} \\ 2c+m+3b &\le 10 &\text{Quality Control} \\ c,\ m,\ b &\ge 0 \end{aligned}\end{cases}$

Following the simplex algorithm , convert to a system of equations:

$\begin{cases}\begin{array}{ccccccccccccccc} z & - & 20c & - & 15m & - & 25b & & & & & & & = & 0 \\ & & 2c & + & m & + & 2b & + & s_1 & & & & & = & 8 \\ & & 2c & + & 2m & + & 3b & & & + & s_2 & & & = & 12 \\ & & 2c & + & m & + & 3b & & & & & + & s_3 & = & 10 \\ \end{array}\end{cases} \quad \begin{array}{c} (0) \\ (1) \\ (2) \\ (3) \end{array}$

In augmented matrix form:

$\left[\begin{array}{ccccccc|c} 1 & -20 & -15 & -25 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 2 & 1 & 0 & 0 & 8 \\ 0 & 2 & 2 & 3 & 0 & 1 & 0 & 12 \\ 0 & 2 & 1 & 3 & 0 & 0 & 1 & 10 \end{array}\right] \quad \begin{array}{c} (0) \\ (1) \\ (2) \\ (3) \end{array}$

The current basic solution, $(z,c,m,b,s_1,s_2,s_3),$ is $(0,0,0,0,8,12,10).$ It is not optimal because there are negative coefficients in row $(0).$

Enter $c$ as a basic variable. Observe the ratios for each row:

$\begin{array}{cc} (1): & \frac{8}{2}=4 \\ (2): & \frac{12}{2}=6 \\ (3): & \frac{10}{2}=5 \\ \end{array}$

The minimum ratio is in row $(1).$ Eliminate $c$ in all rows except row $(1)$ :

$\left[\begin{array}{ccccccc|c} 1 & 0 & -5 & -5 & 10 & 0 & 0 & 80 \\ 0 & 2 & 1 & 2 & 1 & 0 & 0 & 8 \\ 0 & 0 & 1 & 1 & -1 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & -1 & 0 & 1 & 2 \end{array}\right] \quad \begin{array}{c} (0) \\ (1) \\ (2) \\ (3) \end{array}$

The current basic solution is $(80,4,0,0,0,4,2).$ It is not optimal because there are negative coefficients in row $(0).$

Enter $m$ as a variable. Observe the ratios for each row:

$\begin{array}{cc} (1): & \frac{8}{1}=8 \\ (2): & \frac{4}{1}=4 \\ \end{array}$

The minimum ratio is in row $(2).$ Eliminate $m$ in all rows except row $(2)$ :

$\left[\begin{array}{ccccccc|c} 1 & 0 & 0 & 0 & 5 & 5 & 0 & 100 \\ 0 & 2 & 0 & 1 & 2 & -1 & 0 & 4 \\ 0 & 0 & 1 & 1 & -1 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & -1 & 0 & 1 & 2 \end{array}\right] \quad \begin{array}{c} (0) \\ (1) \\ (2) \\ (3) \end{array}$

Note that if this solution is feasible, then it is optimal, because all coefficients in row $(0)$ are positive. Eliminate $b$ in all rows except $(3)$ to find the solution:

$\left[\begin{array}{ccccccc|c} 1 & 0 & 0 & 0 & 5 & 5 & 0 & 100 \\ 0 & 2 & 0 & 0 & 3 & -1 & -1 & 2 \\ 0 & 0 & 1 & 0 & 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 1 & -1 & 0 & 1 & 2 \end{array}\right] \quad \begin{array}{c} (0) \\ (1) \\ (2) \\ (3) \end{array}$

This gives the solution $(100,1,2,2,0,0,0).$ Thus, the factory's maximum profit is $\boxed{\$100}.$

Edit : There is more than one optimal solution. There exists an edge of the feasible region (the intersection of the 1st and 2nd equations) that is parallel to the objective function. Therefore, every point on that edge is also an optimal solution. An integer solution is $(100,2,4,0,0,0,0)$ (2 cars, 4 motorcycles, 0 boats). Note that the maximum profit is the same.

Let

$a=$ the number of toy cars produced

$b=$ the number of toy motorcycles produced

$c=$ the number of toy boats produces

To maximize the profit, we should use all the workers : casting, which has $8$ workers; assembly, which has $12$ workers; quality control, which has $10$ workers.

For casting, which has $8$ workers, we have

$2a + b + 2c = 8 \ldots \ldots \ldots \ldots (1)$

For assembly, which has $12$ workers, we have

$2a + 2b + 3c = 12 \ldots \ldots \ldots \ldots (2)$

For quality control, which has $10$ workers, we have

$2a + b + 3c = 10 \ldots \ldots \ldots \ldots (3)$

Solve for $a$ , $b$ and $c$

$(3) - (1)$ we have

$c = 2$

$(2) - (3)$ we have

$b = 2$

For $b = 2$ and $c = 2$ we have

$a = 1$

The number of toy motorcycles produced to maximize the profit $= \boxed{2}$