A square and a equilateral triangle are to be formed out of the same piece of wire. The wire is 6 inches long. How do you maximize the total area the square and the triangle contain?

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A square and a equilateral triangle are to be formed out of the same piece of wire. The wire is 6 inches long. How do you maximize the total area the square and the triangle contain?

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Answer 1 · 2016-05-13T11:07:31

$\frac{4 \sqrt{3} L}{9 + 4 \sqrt{3}}$ $(4sqrt(3)L)/(9+4sqrt(3))$ for the square
$\frac{9 L}{9 + 4 \sqrt{3}}$ $(9L)/(9+4sqrt(3))$ for the equilateral triangle
Here $L = 6$ $L = 6$

Explanation:

Let be $L = s + t$ $L = s + t$ the total length as the addition of $s$ $s$ the length used by the square and $t$ $t$ the length used by the triangle.
The square area is $a_{s} = {(\frac{s}{4})}^{2}$ $a_s = (s/4)^2$ and the equilateral triangle area is given by $a_{t} = (\frac{t}{6}) \sqrt{{(\frac{t}{3})}^{2} - {(\frac{t}{6})}^{2}} = \frac{t^{2}}{12 \sqrt{3}}$ $a_t = (t/6)sqrt((t/3)^2-(t/6)^2) = t^2/(12 sqrt(3))$
the total area is then $a = a_{s} + a_{t} = \frac{s^{2}}{16} + \frac{t^{2}}{12 \sqrt{3}}$ $a = a_s+a_t = s^2/16+t^2/(12sqrt(3))$
but $t = L - s$ $t = L-s$ then $a = \frac{{(L - s)}^{2}}{12 \sqrt{3}} + \frac{s^{2}}{16}$ $a = (L-s)^2/(12sqrt(3))+s^2/16$
The area critical point is determined doing $\frac{d a}{d s} = 0$ $(da)/(ds) = 0$ and obtaining $s = \frac{4 \sqrt{3} L}{9 + 4 \sqrt{3}}$ $s = (4sqrt(3)L)/(9+4sqrt(3))$ and also $t = \frac{9 L}{9 + 4 \sqrt{3}}$ $t = (9L)/(9+4sqrt(3))$

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A square and a equilateral triangle are to be formed out of the same piece of wire. The wire is 6 inches long. How do you maximize the total area the square and the triangle contain?

1 Answer

Explanation:

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