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A Farmer wants to a fence a rectangular region of $600m^2$ and then divide the middle of it also with a piece of fence that is parallel to one of the sides of the rectangle. What are the dimensions of the region that minimizes the quantity of fence?

I did the following:

$$xy+y=600+y \\ xy=600 \\y=\frac{600}{x}$$

Then, the derivative of this would be:

$$f'(x)=\frac{-600}{x^2}$$

I guess I should be able to find $f'(x)=0$, but this function has no zero. I probably did something brutally wrong, but I have no idea what.

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  • $\begingroup$ If $x$ is the dimension corresponding to three parallel "sides", then the amount of fence used is $P=3x+2y$, $\endgroup$ – David Mitra Jun 3 '15 at 13:55
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Suppose $x$ and $y$ are the two dimensions, with the separating fence measuring $y$.

Since the area is $xy$ we want $xy=600$ or $y=600/x$

What we want to minimize is the fence used, in this case $2x+3y=2x+3\frac{600}{x}=2x+1800x^{-1}$.

Lets find the zeros by differentiating, we get: $2-1800x^{-2}=0\iff 1800x^{-2}=2\iff x^2=900\iff x=30$

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Let $x$ and $y$ be the lengths of the rectangle. Assuming that we have a fence in the middle parallel to the side $y$, we want to minimize $$3y + 2x \; , \quad \text{such that } xy = 600 \text{ holds} \; .$$ Since $x$ and $y$ can not be zero, we can write $$ x = 600 / y \; , $$ and put this into $3y + 2x$, so we get $$ f(y) := 3y + 2 \frac{600}{y} = 3y + \frac{1200}{y} \; .$$ This function $f$ we want to minize, so by calculating the derivative of $f$ and setting it to zero, we have $$f'(y) = 3 - 1200 \frac{1}{y^2} \stackrel{!}{=} 0 \; .$$ Solving this equation gives $$ 3y^2 = 1200 \quad \Leftrightarrow \quad y^2 = 400 \; ,$$ which means that $y = 20$, because $y < 0$ does not make any sense here. You should check, that $y = 20$ is indeed a point, where $f$ attains a mimimum! Now we get $x$ by $$ x = \frac{600}{20} = 30 \; .$$

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Hint: $A = 600 = xy$

The amount of fencing, which we wish to minimize, is $$3x + 2y$$

You should probably draw a rectangle and join two opposite sides at their midpoints to see the truth of the above.

So we use the area to solve for one variable in terms of the other, substitute that into the optimization equation, take the derivative, etc.

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  • $\begingroup$ I left off a "y" - too many questions about using a barn or a river on one side! $\endgroup$ – The Chaz 2.0 Jun 3 '15 at 14:41

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