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I've gotten stuck on my homework with a word problem that asks to find two equations for cost. I'm not really sure where to start, exactly! Word problems are my downfall every single time without fail... The question is:

A cylindrical can is to be constructed to hold $400 \text{ cm}^3$ of liquid. The cost of the material for the circular top and bottom of the can is $3$ cents per square centimetre and the cost of the material used for the curved sides is $2$ cents per square centimetre. Find two equations for the cost, $C$, in terms of i) the radius of the top and bottom, $r$, and ii) the height of the can, $h$.

(Does the area of a circle come into play in finding the answer?)

If someone could walk me through this, that would be great! Thanks in advance!

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Tell us what you have tried for the area of the circles and of the other part (which can be unrolled into a rectangle) and then for the costs. –  Henry Oct 6 '11 at 6:43
    
Is the information about the volume irrelevant? Is it possible to just say C(r) = 0.03r^2 ? 0.03 being the cost, r^2 being the radius of both the top and bottom? Would that be correct or am I totally wrong? –  UVic Student Oct 6 '11 at 7:01
    
No, the information about the volume is needed to get a relationship between $r$ and $h$. You can’t say $C(r)=0.03r^2$; you can’t even use the correct formula for the area of a circle and say that $C(r)=2(0.03)\pi r^2$, since that gives only the cost of the top and bottom (the $2$ is to include both top and bottom); it ignores the side of the can. –  Brian M. Scott Oct 6 '11 at 7:04
    
The volume helps you find the height given the radius and vice versa as pedja has shown. On your cost: the area of a circle is $\pi r^2$ which you need to double as there are two ends, and add the cost of the other part (and watch out for whether you are answering in dollars or cents) –  Henry Oct 6 '11 at 7:07

2 Answers 2

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Work out the total cost in terms of the dimensions $r$ and $h$. The area of a circle of radius $r$ is $\pi r^2$, and we have to count the top and bottom separately, so the total cost of the top and bottom is $2(0.03)\pi r^2$. The circumference of the can is $2\pi r$, and the height is $h$, so the area of the side is $2\pi rh$, and the cost of the side is $2(0.02)\pi rh$. The total cost is therefore $$C = 0.06\pi r^2 + 0.04\pi rh.\tag{1}$$ In order to express this in terms of just $r$ or just $h$, you must find some relationship between $r$ and $h$ that lets you solve for one of them in terms of the other. This is where the known volume comes in. The volume is the area of the base times the height: $V = \pi r^2h$. But we know what this is: it’s $400$, so $\pi r^2h = 400$. Now you can solve this for $h$ in terms of $r$ and substitute into $(1)$ to get $C$ in terms of $r$, and you can also solve it for $r$ in terms of $h$ and substitute into $(1)$ to get $C$ in terms of $h$.

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Thank you so much! Makes complete sense, now I feel silly! –  UVic Student Oct 6 '11 at 7:29

$C=2r^2\pi C_1+2r\pi hC_2$

$C=2r\pi (rC_1+hC_2)$ ,and we know that $V=r^2\pi h \Rightarrow$$ h=\frac{V}{r^2 \pi}$ ,or $r=\sqrt{\frac{V}{\pi h}}$

Now you have to substitute $h$ in equation for $C$ to get the first equation in terms of $r$ and after that make substitution of $r$ in equation for $C$ in order to get second equation in terms of $h$

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