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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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    @Hautdesert, my comment related to the question before the edit which says the top of the diagram is $2x$.2011-10-15

2 Answers 2

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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.2011-10-15
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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