Hey I've been struggling with the algebraic manipulation aspects of this question. I would love any help
A solid cylinder, radius r and height h, is to be constructed under the condition that the sum of its height and circumference is S, where S is constant.
a) if it is given its maximum possible volume, h=S/3
b) if it is given its maximum possible surface area, r+h = S/2
\[ \text{Note that the condition we're given is}\\ \boxed{h + 2\pi r = S}.\]
And keep in mind that \(S\) is constant, so we do not need to worry about it until after computing the first derivative.
\[ \text{The volume of the cylinder is }V = \pi r^2 h.\\ \text{Since }h = S - 2\pi r\text{ we have}\\ V = \pi r^2 (S - 2\pi r) \implies \boxed{V = \pi (Sr^2 - 2\pi r^3) } \]
\[ \text{The surface area of the cylinder is }A = 2\pi r^2 + 2\pi r h.\\ \text{Since }h = S - 2\pi r\text{ we have}\\ A = 2\pi r^2 + 2\pi r (S - 2\pi r) \implies \boxed{A = (2\pi - 4\pi^2)r^2 + 2\pi S r} \]
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\[ \text{For each of these, we will need to use }\boxed{h + 2\pi r = S}\text{ a few times.} \]
\[ \text{Setting }\frac{dV}{dr} = 0\text{ gives}\\ \begin{align*} \pi (2Sr - 6\pi r^2) &= 0\\ Sr - 3\pi r^2 &= 0\\ S - 3\pi r &=0 \tag{**}\\ S &= 3\pi r\\ S&= \frac32(S-h)\\ 0 &= \frac12 S - \frac32 h\\ 3h&=S\\ h&=\frac{S}{3} \end{align*} \]
(**) - You actually have \(r=0\) or \(S - 3\pi r = 0\). To do this correctly, you would then look at the second derivative to check that \(r=0\) actually minimises the volume (this makes sense, because if the radius is 0 then the volume should also be 0). This is a much safer way of discarding the solution \(r=0\), than to just divide it out like I did so there.
\[ \text{Setting }\frac{dA}{dr} = 0\text{ gives}\\ \begin{align*} 2(2\pi - 4\pi^2) r + 2\pi S &=0\\ (2\pi - 4\pi^2 r) + \pi (h + 2\pi r) &= 0\\ 2\pi r - 2\pi^2r^2 + \pi h &= 0\\ 2r - 2\pi r + h &= 0\\ 2r + (h-S) + h &= 0\\ 2(r+h) &= S\\ r+h &= \frac{S}{2} \end{align*}\]
Testing for the max with the second derivative is left as your exercise.