# Maximizing the Volume of Body under a Function

Given is the function $y = 1/2 (4-x)\sqrt{x}$

One has to calculate a) the volume of revolution between function and x-axis, restricted by the function's zeros. b) What is the biggest volume possible of a cone inscribed inside of this revolution volume and the cone's tip at (0,0)?

For a) I have calculate the following: $\pi \int_{a}^b (f(x))^2 dx$ , i.e. $$\pi\int_{0}^4\left(4x - 2x^2 + \frac{x^3}4\right)dx = \pi \left[2x^2 - \frac23 x^3 + \frac{x^4}{16}\right]_{0}^4 = 16.6$$ Is this correct?

Help for b) would be appreciated

Sketch the curve. Suppose that the height of the cone is $x$. So the centre of the base of the cone is at the point $(x,0)$. Then the radius of the base is $\frac{1}{2}(4-x)\sqrt{x}$.
It follows from the usual formula for the volume of a cone ("$\frac{\pi}{3}r^2h$") that the volume of the cone is $$\frac{\pi}{3}\left(\frac{1}{4}(4-x)^2 x\right)x$$ We want to maximize this, with the restriction that $0\le x\le 4$. Equivalently, we want to maximize $(4-x)^2 x^2$. Equivalently, we want to maximize $x(4-x)$. And this is not difficult, with or without calculus.
Remark: The volume of the solid of revolution was computed correctly. I get $16.75516$. This is far enough from your answer that I suspect you are rounding prematurely. Use the "memory" feature of your calculator.
• The apex of the cone is at $(0,0)$. so the cone, assumed right-circular, has axis along the $x$-axis. The only question is the optimal place $(x,0)$ for the centre of the base to be. And on another subject, please note that although your method for (a) is correct, there is a little numerical glitch. – André Nicolas Mar 2 '13 at 8:16
• You are welcome. And to answer your first question slightly more fully, the radius of the base is the $y$ that corresponds to the $x$. A sketch will let you see what is going on. The function increases then decreases, so there is no issue of the cone "cutting through" the solid of revolution. – André Nicolas Mar 2 '13 at 8:38