Let the mixture contain x toys of type A and y toys of type B. Clearly, x, y ≥ 0. Let us construct the following table from the given data:

$\begin{array}{lcccccc} \textbf{Type of Toys} & \textbf{No. of Toys} & \textbf{Machine I time} &\textbf{Machine II time} &\textbf{Machine III time} &\textbf{Profit} \\ & & (\text{in min}) & (\text{in min})& (\text{in min}) & \text{Rs.}\\ A & x & 12x&18x&6x&7.5x\\ B&y&6y&0&9y&5y\\ \text{Total} & x+y& 12x+6y&18x&6x+9y&7.5x+5\\ \textbf{Requirements}& & \text{Max}\; 6\times60=360&\text{Max}\; 6\times60=360&\text{Max}\; 6\times60=360&\\ \end{array}$

We need to maximize Z =7.5x+5y given the following constraints:

$(1):\qquad$12x+6y$\;\leq\;$360 $\to$ 2x+y $\leq\;$ 60

$(2):\qquad$18x $\leq\;$ 360 $\to$ x $\leq\;$ 20

$(3):\qquad$6x+9y$\leq\;$ 360 $\to$ 2x+3y $\leq\;$ 120

$(4):\qquad$ x $\geq$ 0 and $(5):\qquad$ y $\geq$ 0.

$\textbf{Plotting the constraints}$

First draw the graph of the line 2x+y=60.

If x = 0 $\to$ y=60, and if y=0 $\to$ 2x = 60 $\to$ x = 30.

At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 10. So the area associated with this inequality is bounded and towards the origin

Next, draw the line x = 20.

At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 20. So the area associated with this inequality is bounded and towards the origin

Next, plot the line 2x+3y=120.

If x = 0 $\to$ 3y=120 $\to$ y = 40, and if y = 0 $\to$ 2x=120 $\to$ x = 60.

At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 120. So the area associated with this inequality is bounded and towards the origin

$\textbf{Finding the feasible region}$

Since x and y are $\geq$ 0, the feasible region is in the first quadrant.

On solving equations 2x+y = 60 and x = 20, we get:

Substituting for x=20, 2 $\times$ 20 + y = 60 $\to$ y = 60-40 = 20.

$\to$ Point G (20,20)

On solving equations 2x+y=60 and 2x+3y = 120, we get:

Substituting for 2x=60-y, we get 60-y+3y=120 $\to$ 2y = 60 $\to$ y = 30.

Therefore, 2x = 60 - 30 = 30 $\to$ x=15.

$\to$ Point H (15,30)

On solving equations x=20 and 2x+3y = 120, we get:

Substituting for x=20 $\to$ 2 $\times$ 20 + 3y=120 $\to$ 3y = 120-40 = 80 $\to$ y = $\large\frac{80}{3}$

$\to$ Point I (20,$\large\frac{80}{3}$).

Therefore the feasible region is the area bounded by the vertex corner points OEGHCO.

$\textbf{Solving the objective function using the corner point method}$

The values of Z at the corner points are calculated as follows:

$\begin{array}{cc} \textbf{Corner Point} & \textbf{ Z = 7.5x+5y} \\ (0,0) & 0\\ (20,0) & 150 \\ (20,20) & 250 \; \\ (15,30) & 712.5 \; \textbf{(Max Value)} \\ (0,40) & 200 \end{array}$

$\textbf{A) Maximum value of Z is 712.5.}$

It takes $15$ toys of type $A$ and $30$ toys of type $B$ to maximize the profits to $Rs. 712.50$.