- #1
FaraDazed
- 347
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Homework Statement
a cylindrical tin can with volume 0.3l is being made, with the top and bottom sufaces twice the thickness as the sides.
Show that a height to radius ration of [itex]h=4r[/itex] will minimise the amount of aluminium required.
Homework Equations
[itex]V=\pi r^2 h \\
A = 2 \pi r^2 + 2 \pi r h
[/itex]
The Attempt at a Solution
I have kind of done it, but I was not sure how to factor in the fact that the top and bottom is twice the thickness of the sides. But my answer is a half of what it should be, or rather I am getting a ratio of [itex]h=2r[/itex] , so I know why my answer is wrong, but would appreciate a little help on how to do it correctly.
First what I did was, using the first equation in the relevant equations section, solved for height and then subsituted it into the equation for surface area and got...
[itex]
A=2 \pi r^2+ \frac{2 \pi r V}{\pi r^2} = 2 \pi r^2+ \frac{2 V}{\pi r}
[/itex]
Then took the derivative with respect to the radius and set to zero...
[itex]
\frac{dA}{dr}=4 \pi r - \frac{2V}{r^2}=0
[/itex]
Then I multiplied through by r^2
[itex]
0=4 \pi r^3 - 2V \\
r=\sqrt[3]{\frac{2V}{4 \pi}} = \sqrt[3]{\frac{V}{2 \pi}}
[/itex]
Then found the radius
[itex]
r= \sqrt[3]{\frac{0.3}{2 \pi}}=0.3628m
[/itex]
Then with that the height
[itex]
h=\frac{V}{\pi r^2}=\frac{0.3}{\pi \times 0.3628^2} = 0.7255m
[/itex]
And one can see that my height is twice the radius rather than four times. And I know that the fact the top and bottom are twice as thick is why, so could I just leave it as it is and say that, or there is probably a much easier way to factor in the twice thickness thing before hand to get the correct result so would appreciate a little help/advice. Thanks :)