Q15. - Algebra 2

Question

Graph each system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the given function.

15. How many soccer balls and volleyballs should be made to maximize the profit?

Step-by-Step Solution

Answer

To get maximum profit of $55050,5250$soccer balls and $7200$ volleyballs 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 soccer balls and volleyballs produced respectively.

Step-3 –Determination of maximize profit

Profit of $1$soccer ball $=\$5.$

Profit of $x$soccer ball $=\$5x.$

Profit of $1$soccer ball $=\$4.$

Profit of $y$soccer ball $=\$4x.$

Maximize profit function, $f=5x+4y.$

Step-4 –Determination of constraints

Cutting requires $2$hours to make $75$soccer balls and $3$hours to make $60$volleyballs.

Cutting has $500$hours available.

Time for cutting $75$soccer balls =$2$hours.

Time for cutting $1$soccer ball $=\frac{2}{75}$hours.

Time for cutting  soccer ball of $x$soccer balls$=\frac{2x}{75}$ hours.

Time for cutting $60$volleyballs =$3$hours.

 

Time for cutting $1$volleyball$=\frac{30}{60}$

Time for cutting $y$ volleyballs$=\frac{3y}{60}$ hours.

So, $\frac{2x}{75}+\frac{3y}{60}\le 500.$

Sewing needs$3$ hours to make $75$soccer balls and$2$ hours to make$60$ volleyballs.

Sewing has$450$ hours available.

Time for cutting $75$soccer balls = $3$hours.

Time for sewing $1$soccer ball$=\frac{3x}{75}$ hours.

Time for sewing soccer ball of $x$soccer balls=$2$hours.

Time for sewing $60$volleyballs = $2$hours

Time for sewing  $y$ volleyballs$=\frac{2y}{60}$ hours.

So, $\frac{2x}{75}+\frac{2y}{60}\le 450.$

Subject to the constraints, 

$\begin{array}{l}\frac{2x}{75}+\frac{3y}{60}\le 500.\\ \frac{3x}{75}+\frac{2y}{60}\le 450.\\ x\ge 0,y\ge 0.\end{array}$

Step-5 –Evaluation of solution

The inequalities are,

$\begin{array}{l}\frac{2x}{75}+\frac{3y}{60}\le 500.\\ \frac{3x}{75}+\frac{2y}{60}\le 450.\\ x\ge 0,y\ge 0.\end{array}$

The linear equations of the inequalities are
$\begin{array}{l}\frac{2x}{75}+\frac{3y}{60}=500.\\ \frac{3x}{75}+\frac{2y}{60}=450.\\ x=0,y=0.\end{array}$

The points which satisfy the equation $\frac{2x}{75}+\frac{3y}{60}=500$are $\left(0,10000\right)$and $\left(18750,0\right)$.

The points which satisfy the equation $\frac{3x}{75}+\frac{2y}{60}=450$are $\left(0,13500\right)$ $\left(1250,0\right)$.

Step-6–Shading the region

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

The point$\left(0,0\right)$ satisfies the inequalities $\frac{2x}{75}+\frac{3y}{60}\le 500$, $\frac{3x}{75}+\frac{2y}{60}\le 450$ $x\le 0,y\le 0.$

Step-7–Plotting the graph

So, the graph of the inequalities is,

The common shaded region is $OABC$are$O\left(0,0\right),A\left(11250,0\right),B\left(5250,7200\right),C\left(0,10000\right)$

Step-8–Evaluating the maximum and minimum value

The maximization function is $f=5x+4y.$

At point $O\left(0,0\right),$

$f=0.$

At point $A\left(11250,0\right)$

$f=56250.$

At point $B\left(5250,7200\right)$

$f=55050.$

At point $C\left(0,10000\right)$

$f=40000.$