Cylinder inscribed in a sphere [Pre-Calculus] A cylinder is inscribed in a sphere with a radius of 20cm. Find the height, radius, and volume of the largest possible such cylinder.
Radius of cylinder is $R$, radius of sphere is $r$, height is $h$, and volume is $V$.
I used Pythagorean's Theorem to get $R^2 + \dfrac{h^2}{4} = r^2$. I then substituted $R^2 = 400 - \dfrac{h^2}{4}$ into the volume formula for a cylinder $V=\pi R^2h$.
Now I have a equation, but I am completely lost as to how to get any other variable. I am trying to get the maximum by graphing, can not figure that out either. 
 A: You know the formula for the volume of a cylinder: $V=\pi r^2 h$. You can express the radius of the cylinder $r$ in terms of its hight $h$ and one known quantity which in this case is the radius of the sphere that's equal to $20$ cm: $r^2=20^2 - \left(\frac{h}{2}\right)^2$. Substitute $r^2$ in the formula with this expression and you get the following: $V=\pi h\left(400 - \frac {h^2}{4}\right)$. After some fairly basic algebraic manipulations, what you've got now is nothing but a function of one independent variable with respect to $h$: $V(h)=400\pi h - \frac{\pi h^3}{4}$. Your task is to find the maximum value among all possible values this function yields as you plug in different values for $h$ that are greater than zero (negative $h$ values have no meaning here since hight can't be a negative number nor can it be a zero). The easiest way to do this is to take the first derivative of this function with respect to $h$, set it equal to zero and solve the resulting quadratic equation for $h$. The idea here is that as the curve moves up and down along the x-axis, the slop of the line tangent to the graph assumes different numerical values and when the slope is zero there is something called an inflection point where the curve makes a turn. That fact can be used to analyze the behavior of a function in regards to things such as maximum and minimum values that the function reaches at particular points. And that's just the rough idea of what we're going to do here.
$$
V'(h)=\left(400\pi h - \frac{\pi h^3}{4}\right)'= 400\pi - \frac {3\pi h^2}{4}
$$
That's the first derivative and now set it equal to zero and solve for h.
$$
400\pi - \frac {3\pi h^2}{4} = 0\\
h _{1,2}=\pm\frac{40}{\sqrt{3}}
$$
You get two roots, of course, but you're really interested only in the positive one for the reasons mentioned above and that's going to be the hight that gives you the maximum volume of your inscribed cylinder in a sphere of radius $20$ cm: 
$$
h=\frac{40}{\sqrt{3}}
$$
So, you now have to found out what the radius of your cylinder is. That's easy to do. You've already got the expression for the radius. Just plug your $h$ value in:
$$
r=\sqrt{400 - \left(\frac{1}{2} \cdot \frac{40}{\sqrt{3}}\right)^2}=20\sqrt{\frac{2}{3}}
$$
To find the volume of the cylinder, just plug in $h$ and $r$ that you've found into the volume formula:
$$
V=\pi \left(20\sqrt{\frac{2}{3}}\right)^2 \frac{40}{\sqrt{3}} = \pi\frac{32,000}{3\sqrt{3}}
$$
Without calculus, however, the only way that I can think of to solve this problem is to plot the graph of the volume function we developed above on a graphing utility such as a graphing calculator and trace the value where the function reaches its maximum on the interval $(0,40)$. But that can only give you an approximation and is not mathematically speaking a very solid solution at all.
A: You have $$R^2+\frac{h^2}{4}=r^2 \implies R^2=r^2-\frac{h^2}{4}.$$
Secondly, $$V = \pi R^2 h = \pi h\left(r^2-\frac{h^2}{4}\right).$$
It can be shown that $\pi h\left(r^2-\frac{h^2}{4}\right)$ achieves its maximum when $h=\sqrt{\frac{4}{3}}r$. (This is easy to prove by taking derivatives, but you can verify this by plotting a graph of $V$ against $h$ for $r=20$). Lastly, $R^2=r^2-\frac{h^2}{4} \implies R = ?$
A: The equation is 
$$V=\pi R^2h=\pi(400 - \dfrac{h^2}{4})h$$
Now using derivatives you differentiate the volume with respect to $h$ and make that equal to 0
$$\frac{dV}{dh}=\pi(400-\frac{3h^2}{4})$$
Can you take this from here. 
