Answer
(a) $A(x)=-x^3+16x$
(b) $(0,4)$
(c) Refer to teh graph below. Largest area happens when $x\approx2.31$
Work Step by Step
Given: $y=16-x^2$
(a) The area $A$ of a rectangle is given by the formula $A=lw$ where $l$ is the length and $w$ is the width.
Keeping in mind that in the point $(x, y)$, $x$ represents the distance of the point from the $y$-axis while $y$ represents the distance of the point from the $x$-axis, the figure shows that the rectangle has a width of $x$ and a length (in this case, height) of $y$.
Thus, the area of the rectangle is:
$$\begin{align*}
A&=xy\\
A&=x(16-x^2)\\
A&=16x-x^3\\
A&=-x^3+16x
\end{align*}$$
Therefore, $A(x)=-x^3+16x$.
(b) Since $(x,y)$ is in the first quadrant, the domain of $A$ is $x\in(0,4)$.
(c) Use a graphing utility to graph $A(x)=-x^2+16x$. Since the graph represents the area of the rectangle, the value of $x$ that gives the largest Identify the $x$-coordinate of the point see graph, the area is largest area is the $x$-coordinate of the highest point.
Thus, $x\approx2.31$
