Question #1b8d8

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Question #1b8d8

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Answer 1 · 2017-04-17T13:40:13

See below.

Explanation:

$L = l_{s} + l_{e}$ $L = l_s+l_e$ where

$l_{s}$ $l_s$ is the piece associated to the square.
$l_{e}$ $l_e$ is the piece associated to the equilateral triangle.

The associated areas are:

$A_{s} = {(\frac{l_{s}}{4})}^{2} = \frac{l_{s}^{2}}{16}$ $A_s = (l_s/4)^2 = l_s^2/16$ is the square area

$A_{e} = \frac{1}{2} (\frac{l_{e}}{3}) (\frac{l_{e}}{3}) \frac{\sqrt{3}}{2} = l_{e}^{2} \frac{\sqrt{3}}{36}$ $A_e = 1/2(l_e/3)(l_e/3)sqrt(3)/2 = l_e^2sqrt(3)/36$

so the total area is

$A = A_{s} + A_{e}$ $A = A_s+A_e$

so substituting for $l_{e} = L - l_{s}$ $l_e = L - l_s$

$A = \frac{l_{s}^{2}}{16} + {(L - l_{s})}_{s}^{2} \frac{\sqrt{3}}{36}$ $A = l_s^2/16+(L-l_s)^2sqrt(3)/36$

The maximum is attained at $l_{s} = l_{s}^{0}$ $l_s=l_s^0$ such that

$\frac{d A}{d l_{s}}]_{l_{s} = l_{s}^{0}} = 2 \frac{l_{s}^{0}}{16} - 2 (L - l_{s}^{0}) \frac{\sqrt{3}}{36} = 0$ $(dA)/(dl_s)]_(l_s=l_s^0) = 2l_s^0/16-2(L-l_s^0)sqrt(3)/36 = 0$

Solving for $l_{s}$ $l_s$ we get

$l_{s} = \frac{4 \sqrt{3} L}{9 + 4 \sqrt{3}} \approx 4.34965$ $l_s = (4 sqrt[3] L)/(9 + 4 sqrt[3]) approx 4.34965$

and

$l_{e} = 10 - 4.34965 = 5.65035$ $l_e=10-4.34965=5.65035$

This solution is a minimum solution because

$\frac{d^{2} A}{d l_{s}^{2}} = \frac{1}{8} + \frac{1}{6 \sqrt{3}} > 0$ $(d^2A)/(dl_s^2)=1/8 + 1/(6 sqrt[3]) > 0$ characterizing a minimum.

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Question #1b8d8

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