I want to build a half-cylinder structure, with the volume $1200$ cubic feet. But I want to find the minimum surface area needed to achieve this volume. I'm struggling.
I know that the equation for the surface area I'll need is
$A=\pi r^2+h\pi r.$
as well as the volume equation
$V=\pi r^2h/2.$
I'd like $r \geq 6.5$ ft.
$\endgroup$ 11 Answer
$\begingroup$The standard approach appears to use calculus, but we can also do it neatly using the AM-GM inequality.
Solving for $h$ using the equation for volume, we get
$$h=\frac{2400}{\pi r^2}$$
and hence
$$A=\pi r^2+\frac{2400}{r}$$
Now, by the AM-GM inequality, we have
$$A=\pi r^2+\frac{1200}{r}+\frac{1200}{r}\ge3\sqrt[3]{\pi r^2\times\frac{1200}{r}\times\frac{1200}{r}}=3\sqrt[3]{1200^2\pi}\approx496$$ with equality when
$$\pi r^2=\frac{1200}{r}=\frac{1200}{r}$$ which occurs at $$r=\sqrt[3]\frac{1200}{\pi}\approx 7.26$$
Hence the minimum surface area of $\approx 496 \text{ ft}^2$ is achieved when $r\approx 7.26\text{ ft}$.
By the way, if you want a calculus solution as well, I can include it since I wrote it up before I realized there was a nice way to go about it.
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