2. A container in the form of a cylinder, covered on the bottom and the top, is made to hold 8π ft3. Find the dimensions of the cylinder tha

 A container in the form of a cylinder, covered on the bottom and the top, is made to hold 8π ft3. Find the dimensions of the cylinder that will require the least amount of material, that is the surface area, to make.

$$\\V=\pi r^2h=8\pi\\\\ r^2h=8\\\\ r^2=\frac{8}{h}\\\\ r=\sqrt{\frac{8}{h}}\\\\ $Let surface area=A$\\\\ A=2\pi rh+2\pi r^2\\\\ A=2\pi h\sqrt{\frac{8}{h}}+2\pi \frac{8}{h}\\\\ $This will be a max or a min when $\frac{dA}{dh}=0\\\\ A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\$$

$$\\A=2\pi h^{0.5}\sqrt{8}+2\pi *8h^{-1}\\\\ A=4\sqrt2\pi h^{0.5}+16\pi h^{-1}\\\\ \frac{dA}{dh}=2\sqrt2\pi h^{-0.5}-16\pi h^{-2}\\\\ \frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\\\$$

$$\\$Find stat points $\frac{dA}{dh}=0\\\\ 2\sqrt2\pi h^{-0.5}-16\pi h^{-2}=0\\\\ \sqrt2 h^{-0.5}-8 h^{-2}=0\\\\ \sqrt2 h^{1.5}-8=0 \qquad $I mult both sides by $ h^2\\\\ h^{1.5}=\frac{8}{\sqrt2}\\\\ h=[\frac{8}{\sqrt2}]^{2/3}\\\\ h=\frac{2^2}{2^{1/3}}\\\\ h=2^{5/3}\\\\$$

$$\frac{d^2A}{dh^2}=-\sqrt2\pi h^{-1.5}+32\pi h^{-3}\\$$

$${\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{1.5}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}\right)}^{\left(-{\mathtt{3}}\right)} = {\mathtt{2.356\: \!194\: \!490\: \!192\: \!344\: \!8}}$$

GOOD    (I  was hoping  this answer would be positive, otherwise there would be an error)

$$If\; h=2^{5/3}\quad $A minimum amount of material will be used$$ $

$$\\r=\sqrt{\frac{8}{h}}\\\\ r=\sqrt{\frac{2^3}{2^{5/3}}}\\\\ r=\sqrt{2^{4/3}}\\\\ r=2^{4/6}\\\\ r=2^{2/3}\\\\ r=\sqrt[3]{4}\\\\$$

So the minimum amount of material will be needed if       $$r=\sqrt[3]{4}\quad and \quad h=\sqrt[3]{32}$$

(The units are in feet)

This answer is identical to Alan's but our methods are a bit different     

Notice that the height is twice the radius (or height is the same as the diameter).

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