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Here is the question:

A rancher has 300 feet of fencing and needs to make three pens for his animals in the following shape:

Fences

a) Write a formula for the total fencing needed.

b) Find a formula using only $x$ for the total area.

c) Find the dimensions for the pens that maximizes the total area.

My problem is that the diagram seems to suggest that $x=2y$, but we cannot assume that, so the answer for a) is an expression involving both $x$ and $y$. But then for b), I can't seem to be able to express y in terms of x without assuming that $x$ does in fact equal $2y$. But then, if that is the case, we can rewrite the expression for a) so that it is in terms of only $x$ and since we have 300 ft. of fencing, then we can find a value for $x$, and there really is no optimization needed, since there is only one possible configuration. Is the problem written incorrectly or do I just need to go back to Calculus I?

EDIT: I forgot to add another piece of information that was on the diagram. The diagram also explicitly states that the top of the diagram is $2x$ because the other length is also $x$:

Fences, again

Does this change anything?

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If the diagram is right, the problem is wrong. But I don't see how you get any answer for a) involving just $x$ and $y$ (if you don't assume $x=2y$). –  Gerry Myerson Oct 15 '11 at 5:09
    
Suppose that $x=2y$. Then you can find an expression for the area that involves only $x$, and there is more than one possible configuration. There are other possibilities more or less consistent with the picture. But it is true that you need to assume something that is not explicitly stated. –  André Nicolas Oct 15 '11 at 5:18
    
There is another relation implied by the diagram that is not stated. It looks like the width of the right hand rectangles is $x$ as well, but that is not specified. –  Ross Millikan Oct 15 '11 at 5:30
    
@Gerry Myerson: If you assume that the lines in the diagram are straight lines, then all vertical lines are of length $2y$, the top and bottom horizontal lines are $2x$ and the small horizontal line on the right side is x which gives $5x+6y$. –  Hautdesert Oct 15 '11 at 6:08
    
@AndréNicolas: How can there be more than one configuration? If I assume that $x=2y$ then I can express $5x+6y$ as $8x$ and then, setting this equal to $300$ we get $x=75/2$. This is because you can construct the diagram only using line segments of $x$. Or perhaps I am wrong? What other configurations are there? –  Hautdesert Oct 15 '11 at 6:13
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2 Answers

up vote 6 down vote accepted

For fun, we solve the problem without using the calculus.

A glance at the (revised) picture shows that $5x+6y=300$. So we want to maximize $4xy$ subject to $5x+6y=300$. Note the identity $$(5x+6y)^2-(5x-6y)^2=120xy.$$ Since $5x+6y=300$, this can be rewritten as $$120xy=(300)^2-(5x-6y)^2.$$ Note that $120xy$, and hence $4xy$, is largest if $(5x-6y)^2$ is as small as possible, namely $0$. So the maximum is reached when $5x=6y$. From $5x+6y=100$ we then get $10x=300$, so $x=30$ and $y=25$.

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Thanks. I don't know why I couldn't figure this out. –  Hautdesert Oct 15 '11 at 18:10
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$a)$

$f(x,y)=5x+6y$

$b)$

$5x+6y=300 \Rightarrow y=\frac{300-5x}{6}$

$A=A_1+A_2+A_3=2xy+xy+xy=4xy$

$A=\frac{4}{6}x(300-5x)\Rightarrow A=\frac{2}{3}(-5x^2+300x)$

$c)$

$A'_x=0$

$A'_x=\frac{2}{3}(-10x+300)=0 \Rightarrow x=30\Rightarrow y=25$

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