Q25. - Algebra 2

Question

A toy manufacturer is introducing two new dolls, My First Baby and My Real Baby, In one hour, the company can produce 8 First Babies or 20 Real Babies. Because of demand, the company produces at least twice as many First Babies as Real Babies The company spends no more than 48 hours per week making these two dolls. The profit on each First Baby is \(3, and the profit on each Real baby is \)7.50. Find the number and types of dolls that should be produced to maximize profit.

Step-by-Step Solution

Answer

The maximize profit is $\$2160.$

The profit is obtained when $320$ and $160$ first babies and real babies dolls are produced.

Step-1 –Concept of using linear programming

The maximum and minimum value of the functions are determined by using linear programming technique.

Step-2 –Assumption of variables

Let us assume that $x$ and $y$ be the number of first baby dolls and real baby dolls produced by the company respectively.

Step-3 –Determination of maximize profit

Profit on first baby doll $=\$3.00$

Total profit on $x$ first baby doll $=\$3x.$

Profit on $1$ real baby doll$=\$7.50.$

Total profit of $y$real baby doll $=\$7.50.$

Maximize profit, $=3x+7.50y.$

Step-4 –Determination of inequalities

In one hour, the company produces $8$ first babies or $20$ real babies and because of demand, the company produces at least twice as many first babies as real babies $x\ge 2y.$

In one hour, the company produces $8$ first babies or $20$ real babies and because of demand, the company produces at least twice as many first babies as real babies.

$1$First baby doll produced at $\frac{1}{8}$hours.

$x$ First baby doll produced at $\frac{x}{8}$hours.

$1$ Real baby doll produced at $\frac{1}{20}$hours.

$y$ Real baby doll produced at $\frac{y}{20}$hours.

So, $\frac{x}{8}+\frac{y}{20}\ge 48.$

Therefore, the inequalities are,

$\begin{array}{c}x\ge 2y\\ \frac{x}{8}+\frac{y}{20}\ge 48\\ \frac{5x+2y}{40}\ge 48\\ 5x+2y\ge 48\times 40\\ 5x+2y\ge 1920\\ x\ge 0,y\ge 0.\end{array}$

Step-5 –Evaluation of solution

$\begin{array}{c}x\ge 2y\\ 5x+2y\ge 1920\\ x\ge 0,y\ge 0\end{array}$

The linear equations of the respective inequalities are,

$\begin{array}{c}x=2y\\ 5x+2y=1920\\ x=0,y=0\end{array}$

The points which satisfy the equation $x=2y$ are $\left(0,0\right)$ and $\left(2,1\right)$

The points, which satisfy the equation $5x+2y=1920$ $\left(384,0\right)$and $\left(0,960\right)$.

Step-6–Shading the region

We choose the point$\left(0,0\right)$.

The point $\left(0,0\right)$ satisfies the inequality $x\ge 2y$ but it does not satisfies the inequality $5x+2y\ge 1920$.

Step-7–Plotting the graph

So, the graph of the inequalities is,

The common shaded region is $ABCD$and it is unbounded where, $A\left(384,0\right)$and $B\left(320,160\right)$.

Step-8–Evaluating the maximum and minimum value

The maximization function is $=3x+7.50y.$

At point , $A\left(384,0\right)$

$$\begin{gathered} f(x,y)=1152+0 \\ = 1152 \end{gathered}$$

At point $B\left(320,160\right)$.

$\begin{array}{c}f\left(x,y\right)=960+1200\\ =2160.\end{array}$