# Word Problem - Finding a Function for Volume

Yet another word problem that I've been having some trouble with. Again, I just don't know where to start! The question is:

A carpenter has been asked to build an open box with a square base. The sides of the box will cost 4 dollars per square metre and the base will cost 5 dollars per square metre. 60 dollars is available to build the box. If the dimensions of the box are x, the side of the base, and y, the height of the box, find a function for the volume in terems of one of the variables.

I just feel like I'm being bombarded with information and I don't really know where to begin (Or what to begin with!)... Word problems never cease to confuse me, unfortunately.

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Take it a piece at a time. What are you trying to find? You want the volume of the box. That’s the area of the base times the height, so $V = x^2y$; the problem with this, of course, is that it contains both $x$ and $y$, and you need to find an expression that contains only one of them. Presumably that’s what the other information in the problem is good for. Let’s see what we can dig out of it.

There’s only one base, so let’s start there. It will cost $5$ dollars a square metre, so its total cost in dollars will be $5$ times its area. Each side of the square base is $x$ metres long, so the area of the base is $x^2$ square metres, and the cost of the base is $5x^2$ dollars. You can use similar reasoning to get an expression for the cost of one side, which you can then multiply by $4$ to get the total cost of the four upright sides. Adding this subtotal to $5x^2$ will give you the total cost of the entire box, though your expression will again contain both $x$ and $y$.

Fortunately, there’s one piece of information that we’ve not used yet: the total cost is $60$ dollars. Set your expression for total cost equal to $60$; that will give you an equation involving both $x$ and $y$. You can solve that equation for one of the variables in terms of the other. (HINT: It’s much easier to solve for one of them than for the other.) Then you can substitute back into the formula $V=x^2y$ to get $V$ in terms of just one of $x$ and $y$.

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Thank you! I think I get it! Just to be sure I do, the subtotal for the sides would be 4(4xy), therefore 16xy? Using that I solved for y, which gave me 60-5x^2/16x, and I substituted that in for the y in the volume equation. Am I doing it right? –  UVic Student Oct 6 '11 at 7:20
@UVic Student: Looks good to me. –  Brian M. Scott Oct 6 '11 at 7:45
The area of the base is $x^2$ square meters (since it is a square with side length of $x$ meters). There are 4 sides to this box, and each side will be a rectangle with a width of $x$ meters and a height of $y$ meters (the height of the box). The volume of the box is $x^2y$, the area of the base times the height (this is just the volume of a rectangular prism).
Thus, there is $x^2$ square meters of base material needed, and $4xy$ square meters of side material needed. Given the prices, this means that the total price of building the box is $$4(x^2)+5(4xy)=4x^2+20xy\;\;\text{ dollars.}$$ Assuming that the problem means that you are supposed to spend exactly 60 dollars on the box, this means that $$4x^2+20xy=60$$ This constraint will let you determine the volume of the box, $x^2y$, even if you only know the value of $x$ (or, if you only know the value of $y$). Suppose I tell you that $x=1$. Then you know that $$4+20y=60$$ $$y=\frac{56}{20}=\frac{14}{5}$$ Thus, the volume of the box in this case is $$x^2y=1^2\cdot\frac{14}{5}\text{ cubic meters}.$$ You can also solve for $x$, if you know $y$, though it is a little trickier. This means that you will also be able to find the value of $x^2y$ in this case.
The problem is asking you to do this in general - given one of the variables, say $x$, solve for the volume of the box you will get, using the fact that you spend exactly 60 dollars.