Chapter 2.6, Problem 7AYU

7. A rectangle has one corner in quadrant I on the graph of $y\text{ = 16 }-x^2$, another at the origin, a third on the positive $y\text{-axis}$, and the fourth on the positive $x\text{-axis}$. See the figure below.

Express the area A of the rectangle as a function of x.

What is the domain of A?

Graph $A\text{ = }A\left(x\right)$. For what value of x is A largest?

To determine

To find:

a. To express the area $A$ of the rectangle as a function of $x$.

Answer to Problem 7AYU

Solution:

a. $x\left(16-x^2\right)$

Explanation of Solution

Given:

$y=16-x^2$

Calculation:

a. To express the area $A$ of the rectangle as a function of $x$:

Area of the given rectangle $=A= xy$.

Since the rectangle has one corner on quadrant I on the graph of $y=16-x^2$.

Substitute $16-x^2$ for $y$. Then

$A\left(x\right)=x\left(16-x^2\right)$.

To determine

To find:

b. To find the domain of $A$.

Answer to Problem 7AYU

Solution:

b. $\left\{x|0<x<4\right\}$

Explanation of Solution

Given:

$y=16-x^2$

Calculation:

b. To find the domain of $A$:

Since, $A\left(x\right)=x\left(16-x^2\right)$.

The domain of $A\left(x\right)=\left\{x|0<x<4\right\}$.

To determine

To find:

c. To graph $A=A\left(x\right)$ and to find the value of $x$ is $A$ largest.

Answer to Problem 7AYU

Solution:

c.$A\text{is largest when}x=2.31$.

Explanation of Solution

Given:

$y=16-x^2$

Calculation:

c. To graph $A=A\left(x\right)$ and to find the value of $x$ is $A$ largest:

From the graph, we can see that, $A$ is largest when $x=2.31$.