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I have a question which asks:

A cylinder shaped can holds $5000cm^3$ of water. Find the dimensions that will minimise the cost of metal in making the can.

What I did was express the height in terms of the radius as:

$$h=\frac{5000}{(\pi)(R^2)}$$

Then I differentiate it to:

$$\frac{(\pi)(R^2)-2R}{((\pi)(R^2))^2}$$

Now, where should I go from here? should I let it equal zero or should I do something else?

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Hint: The amount of metal needed to make the can would consist of the bottom and the lid (both of which are circles) and the cylinder which when unrolled is a rectangle. Thus, you have to formulate your problem as follows:

Objective: Minimize Area of (bottom + lid + cylinder)

Subject to the condition that:

Volume of cylinder = 5000

Note: You have two variables $r$ and $h$. Use the constraint to eliminate one of these variables from the objective function and then use calculus to find the optimum.

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  • $\begingroup$ Thank you that cleared it up perfetly $\endgroup$ – user May 22 '13 at 17:02
  • $\begingroup$ I have run into another problem. I get to the stage $(4 \pi r + r^2) + \frac{r^2 - 10000(2R)}{(r^2)^2} = 0$ From here I am not sure where to go, if I add the fractions together I end up at the end with $r^3+r^4= \frac{5000}{\pi}$ How do I get R $\endgroup$ – user May 22 '13 at 19:55
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To elaborate slightly on @response's answer, you should minimize $2\pi r^2 + 2\pi r h$ (the area of the cilinder) given that $\pi r^2 h = 5000$ (the volume of the cylinder is 5000). So plugging in the constraint in the objective function, we find that you should minimize $2\pi r^2 + \frac{10000}{r}$. Can you take it from here?

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  • $\begingroup$ Thank you, your answer was also very helpful $\endgroup$ – user May 22 '13 at 17:07

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