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Question

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 fluid ounces =473.176 cubic centimeters16 \text{ fluid ounces } = 473.176 \text{ cubic centimeters}

So, the can has a volume 473.176473.176 cm3^3. That is, πr2h=473.176\pi r^2h = 473.176, or h=473.176πr2h = \dfrac{473.176}{\pi r^2}.

We have to minimise the surface area. For radius rr and height hh, the surface area is given by:

S=2πr2+2πrhS = 2 \pi r^2+ 2 \pi rh

Substitute hh in SS:

S=2πr2+2πr×473.176πr2=2πr2+946.352rS = 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:

dSdr=4πr946.352r2\dfrac{dS}{dr} = 4 \pi r - \dfrac{946.352}{r^2}

Putting this equal to 0 and finding the critical points we get r=(946.3524π)1/3(75.30)1/3r= \left( \dfrac{946.352}{4 \pi} \right)^{1/3} \approx (75.30)^{1/3}. Also, h=473.176πr28.4464h = \dfrac{473.176}{\pi r^2} \approx 8.4464. To check for maxima or minima, we find the second derivative:

d2Sdx2=4π+1892.704r3\dfrac{d^2 S}{dx^2} = 4 \pi + \dfrac{1892.704}{r^3}

For r(75.30)1/3r \approx (75.30)^{1/3}, d2Sdx2\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(75.30)1/3r \approx (75.30)^{1/3} and h8.4464h \approx 8.4464.

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