# max/min problem with areas

You are given 30 meter of material which you will cut into two pieces. One piece will form an equilateral triangle, the other a rectangle whose length is three times its width.

Where should you cut if the combined area is to minimized? How could the combined area of these two figures be maximized?

My work:

$A\left(x\right) = \frac{\sqrt{3}}{4}x^2 + \frac{3}{8}\left(30 - 3x \right)^2$ Fn of Area.

Now $A^{\prime}$ is a linear function, so it can only be minimized because $A^{\prime \prime}\left(x\right) > 0$ due to the 2nd derivative test.

How can it be maximized?????

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## 1 Answer

You need to use the fact that your domain is bounded. The value of $x$, which is the length of the side of your triangle, must be at least zero, and can be at most 10 (since there are only 30 meters of material). So the problem is to optimize your function on the interval $0 \le x \le 30$. Find all critical points (there is just one) and evaluate the function at the critical point and the endpoints. The maximum and minimum values will be among the resulting values.

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Just thought it was odd, to use all the material for the triangle to get the max. –  yiyi Nov 6 '12 at 7:58
It is not odd, the rectangle s very unsquarish, it is an inefficient use of material. –  André Nicolas Nov 6 '12 at 8:20
@AndréNicolas so to find efficency of the material, you suggest that the ratio of the perimeter and area is the standard? –  yiyi Nov 7 '12 at 0:37
@MaoYiyi The efficiency of the shape is captured by the ratio of the perimeter squared to the area. This is constant for any particular shape, and the higher this is, the less efficient it is in terms of enclosing areas. An equilateral triangle has a ratio of $36/\sqrt{3}=20.78...$ while the rectangles in your problem have ratio 64/3=21.33... . So, the equilateral triangle is just a little better. –  Matthew Conroy Nov 7 '12 at 1:40
@MatthewConroy Why is the perimeter squared? and is there a name I can search for about this? –  yiyi Nov 7 '12 at 1:45