Chapter 3, Problem 11STP

To determine

To Find: Thearea represented by the intersection of inequalities $y\le 20,x\le 16,y\ge 0,x\ge 0\text{and }y\le -x+32$ .

Answer to Problem 11STP

  $312{\text{ft}}^2$

Explanation of Solution

Given:

Given inequalities are $y\le 20,x\le 16,y\ge 0,x\ge 0\text{and }y\le -x+32$

Calculation:

Plot $y\le 0$ .

Shade the region below x-axis as $y\le 0$ .

Plot $x\ge 0$ .

Shade the region right of the y-axis as $x\ge 0$ .

Plot $y\le 20$ .

Shade the region below a line parallel to x-axis at $y=20$ as $y\le 20$ .

Plot $y\le -x+32$ .

Plot the graph for the coordinates of $y=-x+32$ .

  $\begin{array}{cc}x& y\\ 0& 32\\ 32& 0\end{array}$

Shade the region below the as $y\le -x+32$ .

Plot $x\le 16$ .

Shade the region left of the line parallel to the y-axis at $x=16$ as $x\le 16$ .

Graph is drawn below.

  

Calculating area of the region formed as shown in graph above.

Area $=$ area of rectangle $-$ area of triangle

  $\begin{array}{l}=20\times 16-\frac{1}{2}\times \left(16-12\right)\times \left(20-16\right)\\ =20\times 16-\frac{1}{2}\times 4\times 4\\ =312\end{array}$ .

Therefore, the answer is $312{\text{ft}}^2$ .