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?

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

Let us assume that $x$and $y$be the number of soccer balls and volleyballs produced respectively.

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, $f=5x+4y.$

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,

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

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 $\left(0,13500\right)$ $\left(1250,0\right)$.

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

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

So, the graph of the inequalities is,

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

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

$f=40000.$