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Question:

A fence will create a rectangular area with one side being formed by an existing building (and hence, the fence only needs 3 sides). One side will be created using Redwood fencing and the other two equal sides will be created using Cedar. Redwood fencing costs \$9 per yard and cedar fencing costs $18 per yard. Find the dimensions of the rectangular area with the least cost if the total area enclosed must be 400 square yards. Verify that your answer will produce the minimum cost.

So I have put the domain as $[0,400]$, and I've been trying to use the formula for the area of a rectangle to solve this problem. I figure that I need to disregard the fact that part of the rectangle is a wall for now, and then use that information later to figure out the minimum cost.

This is what I've tried so far (r is redwood fencing and c is cedar fencing): $$A=rc$$ $$400=rc$$ $${400\over r}=c$$ $${400r\over r}-400=0$$ Obviously this turns out to be $0$, and I can't figure out a way to derive this equation to find critical points.

The only thing I can think of is that I do need to take into consideration the fact that part of the rectangle is a wall, but I don't know how to do it.

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    $\begingroup$ How did you produce the last equation from the second-to-last? Also, note that you have disregarded other information: the cost of the fencing, and the fact that two of the sides are cedar while only one is redwood. $\endgroup$ – Barry Smith Jul 14 '15 at 2:15
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    $\begingroup$ Draw a picture. The cost is $9r+(2)(18c)$. But $c=400/r$ so the cost is $9r+\frac{(36)(400)}{r}$. $\endgroup$ – André Nicolas Jul 14 '15 at 2:15
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Let me just re-write the answer that @Andre provided above.

I think you are not optimizing anything. Since you are trying to minimizing the cost in this case, the cost J $$ J = 9r + 2*18*c = 9r + 36c $$

By substituting r = 400/c, you see that the cost becomes: $$ J = \frac{3600}{c} + 36c $$

now that J is a function in one variable, you can differentiate to get $$ \frac{dJ}{dc} = \frac{-3600}{c^2} + 36 $$

setting $ \frac{dJ}{dc} = 0 $, you find that c must be 10. And in this case, r must be 40.

Hope that helped.

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  • $\begingroup$ I think I understand. Is the only use for the area equation to put one variable in terms of the other? If that makes any sense. $\endgroup$ – matryoshka Jul 14 '15 at 2:39
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    $\begingroup$ Essentially, but think of it this way: If the enclosed area is not restricted, you could make it super duper tiny, and the cost of the fencing would be tiny too. That is, the cost $9r + 36c$ can be made very close to zero if you are allowed to choose r and c close to 0. This changes when you restrict the area by the equation rc = 400. If you want to make r very small and maintain this relationship, you must make c very big. You cannot make both small simultaneously. This "constraint" makes determining the minimal cost more difficult (and interesting). $\endgroup$ – Barry Smith Jul 14 '15 at 2:46

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