Page 69 - Calc for the Clueless

The perimeter consists of three sides of the rectangle plus the circumference of the semicircle.

We then solve for w and substitute in the formula for A. We get

Setting dA/dL = 0, we get L = 2p/(4 + π). From the formula for p we get 2w = 2p/(4 + π)!!!!!!!!!!! The
maximum light occurs when the length equals twice the width!!!!?!!!

There are many problems where we are asked to find the maximum surface area or volume of a figure inscribed
in a sphere or other shape. We will look at a messy one.

Example 13—

Find the dimensions of the cone of largest volume that can be inscribed in a sphere of radius R.

First we note the symmetry of the cone inside the sphere. We note that the variables are r and h, the dimensions
of the cone, which are to be found. The only known in the problem is the radius of the sphere R, which is a
given number. In the second picture is the way h, r, and R are related, again a right triangle.

2
2
Since the volume of a cone is (1/3)πr h, it will make a much easier problem to differentiate if we solve for r .
Thus

We can then substitute into the expression for r , and then take the square root to get r.
2