Chapter 5.1, Problem 76STP

a.

To determine

Write a system of inequalities for number of bags that can made.

Answer to Problem 76STP

  $\begin{array}{l}c+3l\le 56\\ 4c+2l\le 104\end{array}$

Explanation of Solution

Given information:

A committee is making canvas tote and leather tote bags with leather handles. For canvas and leather bags they need $4yards$ of canvas and $1yard$ of leather, $3yards$ oh leather and $2yards$ of canvas. The committee purchased $56yards$ of leather and $104yards$ of canvas.

Calculation:

A system of inequalities for number of bags that can made.

Use $c$ for canvas and $l$ for leather bags,

According to the given data,

  $\begin{array}{l}c+3l\le 56\\ 4c+2l\le 104\end{array}$

Hence, system of inequalities for number of bags that can made is,

  $\begin{array}{l}c+3l\le 56\\ 4c+2l\le 104\end{array}$

b.

To determine

Draw a graph for feasible region.

Answer to Problem 76STP

  

Explanation of Solution

Given information:

A committee is making canvas tote and leather tote bags with leather handles. For canvas and leather bags they need $4yards$ of canvas and $1yard$ of leather, $3yards$ oh leather and $2yards$ of canvas. The committee purchased $56yards$ of leather and $104yards$ of canvas.

Calculation:

A graph of system of inequalities for feasible region.

  

Hence, the shaded portion is the feasible region.

c.

To determine

Find the coordinates of vertices for feasible region.

Answer to Problem 76STP

  $\left(0,0\right),\left(20,12\right),\left(26,0\right),\left(0,18\frac{2}{3}\right)$

Explanation of Solution

Given information:

A committee is making canvas tote and leather tote bags with leather handles. For canvas and leather bags they need $4yards$ of canvas and $1yard$ of leather, $3yards$ oh leather and $2yards$ of canvas. The committee purchased $56yards$ of leather and $104yards$ of canvas.

Calculation:

Consider the inequalities,

  $\begin{array}{l}c+3l\le 56\\ 4c+2l\le 104\end{array}$

The equation of inequalities for coordinates of vertices for feasible region,

  $\begin{array}{l}c+3l=56\\ 4c+2l=104\end{array}$

  $\begin{array}{l}putc=0inc+3l=56\\ 0+3l=56\\ l=\frac{56}{3}=18\frac{2}{3}\end{array}$

One coordinate of vertices for feasible region is, $\left(0,18\frac{2}{3}\right)$

  $\begin{array}{l}putl=0in4c+2l=104\\ 4c+2\times 0=104\\ 4c=104\\ c=\frac{104}{4}=26\end{array}$

Second coordinate of vertices for feasible region is, $\left(26,0\right)$

Now put $c=56-3l$ in $4c+2l=104$ equation we get,

  $\begin{array}{l}4\left(56-3l\right)+2l=104\\ 224-12l+2l=104\\ -10l=-120/10\\ l=12\\ now,\\ c=56-3l\\ c=56-3\left(12\right)\\ c=20\end{array}$

Third coordinate of vertices for feasible region is, $\left(20,12\right)$

Fourth coordinate of vertices for feasible region is, $\left(0,0\right)$

Hence, all coordinate of vertices for feasible region are, $\left(0,0\right),\left(20,12\right),\left(26,0\right),\left(0,18\frac{2}{3}\right)$

d.

To determine

Write a function for total profit.

Answer to Problem 76STP

  $f\left(c,l\right)=20c+35l$

Explanation of Solution

Given information:

A committee is making canvas tote and leather tote bags with leather handles. For canvas and leather bags they need $4yards$ of canvas and $1yard$ of leather, $3yards$ oh leather and $2yards$ of canvas. The committee purchased $56yards$ of leather and $104yards$ of canvas.

If they sells canvas and leather bags at a profit of $\$20and\$35$ .

Calculation:

A will be,

  $\begin{array}{l}Assumef\left(c,l\right)\\ f\left(c,l\right)=20c+35l\end{array}$

Hence, function for total profit on both bags is, $f\left(c,l\right)=20c+35l$

e.

To determine

Find the way to make the maximum profit by the club.

Answer to Problem 76STP

  $20$ Canvas bags and $12$ leather bags.

Explanation of Solution

Given information:

A committee is making canvas tote and leather tote bags with leather handles. For canvas and leather bags they need $4yards$ of canvas and $1yard$ of leather, $3yards$ oh leather and $2yards$ of canvas. The committee purchased $56yards$ of leather and $104yards$ of canvas.

If they sells canvas and leather bags at a profit of $\$20and\$35$ .

Calculation:

Consider the function for total profit on both bags,

  $f\left(c,l\right)=20c+35l$

Now find the function value at vertices, $\left(0,0\right),\left(20,12\right),\left(26,0\right),\left(0,18\frac{2}{3}\right)$

The function value at vertices $\left(0,0\right)$

  $\begin{array}{l}f\left(0,0\right)=20\times 0+35\times 0\\ =0\end{array}$

The function value at vertices $\left(20,12\right)$

  $\begin{array}{l}f\left(20,12\right)=20\times 20+35\times 12\\ =820\end{array}$

The function value at vertices $\left(26,0\right)$

  $\begin{array}{l}f\left(26,0\right)=20\times 26+35\times 0\\ =520\end{array}$

The function value at vertices $\left(0,18\frac{2}{3}\right)$

  $\begin{array}{l}f\left(0,18\frac{2}{3}\right)=20\times 0+35\times 18\frac{2}{3}\\ =653.33\end{array}$

Maximum function value is found at $\left(20,12\right)$ .

Hence, maximum profit earned by the club to make $20$ canvas bags and $12$ leather bags.

e.

To determine

Find the maximum profit.

Answer to Problem 76STP

  $\$820$

Explanation of Solution

Given information:

A committee is making canvas tote and leather tote bags with leather handles. For canvas and leather bags they need $4yards$ of canvas and $1yard$ of leather, $3yards$ oh leather and $2yards$ of canvas. The committee purchased $56yards$ of leather and $104yards$ of canvas.

If they sells canvas and leather bags at a profit of $\$20and\$35$ .

Calculation:

Consider the function for total profit on both bags,

  $f\left(c,l\right)=20c+35l$

Now find the function value at vertices, $\left(0,0\right),\left(20,12\right),\left(26,0\right),\left(0,18\frac{2}{3}\right)$

The function value at vertices $\left(0,0\right)$

  $\begin{array}{l}f\left(0,0\right)=20\times 0+35\times 0\\ =0\end{array}$

The function value at vertices $\left(20,12\right)$

  $\begin{array}{l}f\left(20,12\right)=20\times 20+35\times 12\\ =820\end{array}$

The function value at vertices $\left(26,0\right)$

  $\begin{array}{l}f\left(26,0\right)=20\times 26+35\times 0\\ =520\end{array}$

The function value at vertices $\left(0,18\frac{2}{3}\right)$

  $\begin{array}{l}f\left(0,18\frac{2}{3}\right)=20\times 0+35\times 18\frac{2}{3}\\ =653.33\end{array}$

Maximum function value is found at $\left(20,12\right)$ is $\$820$

Hence, maximum profit is $\$820$ .