Question

A rectangle has one vertex in quadrant 1st on the graph of $y=10-x^2$, another at the origin, one on the positive $x$-axis, and one on the positive $y$-axis.

(a) Express the area $A$of the rectangle as a function of $x$.

(b) Find the largest area $A$that can be enclosed by the rectangle.

Short answer

(a) The area of the rectangle as a function of $x$is $10x-x^3$.

(b) The largest area is $12.1716123890037$

Step-by-step solution

Step 1. Part (a) Find the area A of the rectangle as a function of x.

The area of the rectangle is $xy$

Substitute $y=10-x^2$

$$\begin{gathered} A=x\left(10-x^2\right) \\ A=10x-x^3 \end{gathered}$$

Step 2. Part (b) Find the largest area A that can be enclosed by the rectangle.

This is the graph of $A$, so the largest area islocalid="1645957871121" $12.1716123890037$