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# A soda can in the shape of a cylinder is to hold 16 fluid ounces. Find the dimensions of the can that minimize the surface area of the can.

Solution

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First, converting 16 fluid ounces to cubic centimeters.

$16 \text{ fluid ounces } = 473.176 \text{ cubic centimeters}$

So, the can has a volume $473.176$ cm$^3$. That is, $\pi r^2h = 473.176$, or $h = \dfrac{473.176}{\pi r^2}$.

We have to minimise the surface area. For radius $r$ and height $h$, the surface area is given by:

$S = 2 \pi r^2+ 2 \pi rh$

Substitute $h$ in $S$:

$S = 2 \pi r^2+ 2 \pi r \times \dfrac{473.176}{\pi r^2} = 2 \pi r^2 + \dfrac{946.352}{r}$

Differentiating to find the first derivative:

$\dfrac{dS}{dr} = 4 \pi r - \dfrac{946.352}{r^2}$

Putting this equal to 0 and finding the critical points we get $r= \left( \dfrac{946.352}{4 \pi} \right)^{1/3} \approx (75.30)^{1/3}$. Also, $h = \dfrac{473.176}{\pi r^2} \approx 8.4464$. To check for maxima or minima, we find the second derivative:

$\dfrac{d^2 S}{dx^2} = 4 \pi + \dfrac{1892.704}{r^3}$

For $r \approx (75.30)^{1/3}$, $\dfrac{d^2 S}{dx^2}$ is positive. Therefore, it is the point of minimum.

Thus, the surface area is minimised by the dimensions $r \approx (75.30)^{1/3}$ and $h \approx 8.4464$.

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