Of all the closed cylindrical cans (right cir | vedclass

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(A) Let $r$ be the radius and $h$ be the height of the cylinder.Given volume $V = \pi r^2 h = 100 	ext{ cm}^3$.Thus,$h = \frac{100}{\pi r^2}$.The sur

Vedclass · Accepted Answer

Radius $= \left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$,Height $= 2\left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$

Vedclass · Answer

(A) Let $r$ be the radius and $h$ be the height of the cylinder.Given volume $V = \pi r^2 h = 100 	ext{ cm}^3$.Thus,$h = \frac{100}{\pi r^2}$.The surface area $S$ is given by $S = 2\pi r^2 + 2\pi rh$.Substituting $h$,we get $S = 2\pi r^2 + 2\pi r \left(\frac{100}{\pi r^2}ight) = 2\pi r^2 + \frac{200}{r}$.Differentiating with respect to $r$,we get $\frac{dS}{dr} = 4\pi r - \frac{200}{r^2}$.Setting $\frac{dS}{dr} = 0$,we have $4\pi r = \frac{200}{r^2} \Rightarrow r^3 = \frac{50}{\pi} \Rightarrow r = \left(\frac{50}{\pi}ight)^{1/3}$.Checking the second derivative,$\frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3}$. Since $r > 0$,$\frac{d^2S}{dr^2} > 0$,so $S$ is minimum at $r = \left(\frac{50}{\pi}ight)^{1/3}$.Then $h = \frac{100}{\pi (50/\pi)^{2/3}} = \frac{100}{\pi} \cdot \frac{\pi^{2/3}}{50^{2/3}} = 2 \cdot \frac{50}{\pi^{1/3} \cdot 50^{2/3}} = 2 \left(\frac{50}{\pi}ight)^{1/3}$.Thus,the dimensions are $r = \left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$ and $h = 2\left(\frac{50}{\pi}ight)^{1/3} 	ext{ cm}$.

Of all the closed cylindrical cans (right circular) with a given volume of $100 \text{ cm}^3$,find the dimensions of the can that has the minimum surface area.

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