stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.
"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.
So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = \frac{25}{r}-r$$
This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.
Best Answer
$$A_{s} = 2\pi rh + 2\pi r^2 \implies 50\pi = 2\pi rh+2\pi r^2$$
From here, you correctly solved for $h$ in terms of $r$:
$$25\pi = \pi rh+\pi r^2 \implies 25 = rh+r^2 \implies 25 = r(h+r) \implies \color{blue}{h = \frac{25}{r}-r} \tag{1}$$
You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.
$$V = bh = \pi r^2h$$
Plugging in $(1)$, you get
$$V = \pi r^2\left(\frac{25}{r}-r\right) = 25\pi r-\pi r^3$$
Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:
$$\frac{dV}{dr} = 25\pi-3\pi r^2$$
$$\frac{dV}{dr} = 0 \implies 0 = 25\pi-3\pi r^2$$
Can you continue from here?
Edit: You can also use the fact that $\frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -\frac{b}{2a}$ must be a maximum.