Question 975337
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The volume of a cylinder is given by *[tex \Large V\ =\ \pi r^2h], so



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r^2h\ =\ 16]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r^2\ =\ \frac{16}{h}]


The area of either the top or bottom of the can is given by


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \pi r^2],


but there are two ends, so the area of the ends is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2\pi r^2]


The area of the side of the can is the circumference of the end times the height of the can, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2\pi h]


The total surface area of the can is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A\ =\ 2\pi r^2\ +\ 2\pi h]


But if we substitute *[tex \Large r^2\ =\ \frac{16}{h}] we can create a single variable function for the total surface area:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A(h)\ =\ \frac{32pi}{h}\ +\ 2\pi h]


Using the fact that the ends cost twice as much per unit area than the side, we can now write a function for the cost in terms of the height:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  C(h)\ =\ \frac{64pi}{h}\ +\ 2\pi h]


Then take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dC}{dh}\ =\ 2\pi\ -\ \frac{64\pi}{h^2}]


Set the first derivative equal to zero and solve for the value of *[tex \Large h] that yields a local extrema for the function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\pi h^2\ =\ 64\pi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h_{ext}\ =\ 4\sqrt{2}]


Now take the 2nd derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2C}{dh^2}\ =\ \frac{128\pi}{h^3}]


Which is positive for *[tex \Large h\ >\ 0], hence *[tex \Large h_{ext}\ =\ h_{min}]


Now that we have a height value that minimizes the cost function, we can calculate the resulting radius value using the volume relationship.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r^2\ =\ \frac{16}{4\sqrt{2}}\ =\ 2\sqrt{2}\ =\ \sqrt{8}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r\ =\ \sqrt{\sqrt{8}}\ =\ \sqrt[4]{8}]




John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \