### Home > APCALC > Chapter 5 > Lesson 5.1.4 > Problem 5-40

A rectangle is bounded by the function *y* = –*x*^{2} + 5 and the *x*-axis as shown below. Homework Help ✎

If the base of the rectangle is 2

*x*, what is the height?Notice that the rectangle is symmetric across the

*y*-axis.Because of the symmetry, the endpoints of the rectangle can be found at −

*x*and +*x*... after all, the distance between −*x*and*x*is 2*x*, the length of the base.Consequently, the height of the rectangle can be evaluated at

*f*(*x*):

height =*y*= −*x*^{2}+ 5What is the maximum area that the rectangle can enclose?

Area = (base)(height)

Use the base and height you found in part (a).Then optimize the Area function.To find the maximum Area:

1. Let*A*'(*x*) = 0.

2. Solve for*x*.

3. Use the*x*and your expressions for base & height to calculate the maximum area.If you make the rectangle from part (b) into a flag and spin it around the

*x*-axis, what is the resulting volume?A cylinder with height __________ and radius __________.

Determine the value of

*x*so that the volume of the rotated rectangle is a maximum.Volume of a cylinder:

*V*=*πr*^{2}*h*

If you need guidance optimizing the volume, refer to steps in part (b).