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I have a word problem that reads as so

A farmer wants to fence a rectangular part of his land (30,000 square feet). The fenced area is to have one border shared with a neighbor which he wishes to be fancy so he will spend 25 dollars per linear foot on that side where as he will use 5 dollars per linear foot of fencing on the rest. Find the dimensions that will minimize the cost

I set up my equations (x shares the border with the fancy fence)

$ A = xy $

$ C = 25x + 5(x + 2y) = 30x + 10y $

Then I found my x-value

$ x = \frac{30,000}{y} $

Plugged it in

$ C = 30(\frac{30,000}{y}) + 10y = \frac{900,000}{y} + 10y $

Derivative of C

$ C' = \frac{900,000}{y^2} + 10 $

$ -900,000 = -10y^2 $

$ y^2 = 90,000 $

$ y = 300 $

Finally, plugged it into original equation

$ 30,000 = 300x = x = 100 $

I just wanted to check if I've done this right. If not, can you explain where I've gone wrong?

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    Should say you are setting $C'=0$. As I mentioned before, the dimensions you obtained are correct. For completeness, should explain why this gives minimum cost. In many calculus courses, solution would be marked correct without such justification.2012-11-28

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You write:

$C = 30(\frac{30,000}{y}) + 10y = 90,000 = y^2 = y = 300$

Take a good look at that: it says (among other things) that $90,000=300$. Do you really want to say that?

Note too that you’re trying to find an extreme value of something, so you should expect to take a derivative of something.

What you should get from $C=30\left(\frac{30,000}y\right)+10y$ is that $C=\frac{900,000}y+10y$. Now you want to find the value of $y$ that minimizes $C$, so you calculate $\frac{dC}{dy}$, set it to $0$, and solve for critical points. We have

$\frac{dC}{dy}=-\frac{900,000}{y^2}+10\;,$

so we set $-\frac{900,000}{y^2}+10=0$ and solve: $10y^2=900,000$, $y^2=90,000$, and $y=300$. You still have to check that this critical value yields a minimum of $C$, but that’s easy to do: by inspection $\frac{dC}{dy}$ is negative when $y$ is a little less than $300$ and positive when $y$ is greater than $300$.

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    @StrugglingWithMath: What you now have is correct, apart from not including a verification that you’ve actually found the **minimum** cost, but the original version can’t be saved by turning some of the equals signs into implications, because in it you never took a derivative and had no way to conclude that $y^2=90,000$.2012-11-28
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It would be good to explicitly say $x$ is the side shared with the neighbor and $y$ is the other direction. Under "Plugged it in" your equal signs are not all equal. You should have $C=\frac {900,000}y + 10y$, but this is not $90,000, y^2$ or $y$. Now take the derivative with respect to $y$, set it to $0, \ldots$

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    My issue was I should have put arrows instead of equal signs. I separated each step into new lines instead. Does that look better?2012-11-28