Question

Any ideas for this question

Answer 1

Answer:  36pi (choice b)

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Explanation:

Let x = length of side BC

This is the height of the cylinder. Think of a can that is laying on its side. The radius of this can or cylinder is CD

If point B has x coordinate of 12, and BC is 12 units long, this must mean point C has x coordinate of 12-x. This is plugged into the function to show that CD has a length of exactly $\sqrt{12-x}$

This is the radius of the cylinder

The volume of a cylinder is $V = \pi*r^2*h$

Plug in the radius and height mentioned to get this function in terms of x

$V = \pi*\left(\sqrt{12-x} \ \right)^2*x$

That simplifies to

$V = \pi(12-x)x$

or

$V = \pi(12x-x^2)$

Ignore the pi portion for now.

We wish to maximize the function f(x) = 12x-x^2

Use either calculus (specifically derivatives) or a graphing calculator to find that the vertex is at (6, 36)

This means x = 6 leads to the largest f(x) value being 36.

Therefore, the volume V is maxed out when x = 6 and we get a max volume of 36pi cubic units.