# A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.

Second Derivative Test - If f'(x) = 0 at a point and f''(x) > 0 at this point, then f(x) has a local minimum at this point.

Let $s$ be the side of square and $a$ be side of equilateral triangle.

Given,

$4s + 3a = 36$
$a = \frac{36-4s}{3}$  -- ( 1 )

Total area $A = s^2 + \frac{\sqrt{3}}{4}a^2$
Substituting for $a$ from (1) above,

$A = s^2 + \frac{\sqrt{3}}{4} ( \frac{36-4s}{3} )^2$
$A = s^2 + \frac{4\sqrt{3}}{9} (9-s)^2$

$\frac{dA}{ds} = 2s + \frac{4\sqrt{3}}{9}(2s-18)$
$\frac{{d^2}A}{ds^2} = 2 + \frac{8\sqrt{3}}{9} > 0$

Hence based on second derivative test, a minimum happens for A wherever $\frac{dA}{ds} = 0$

So,

$\frac{dA}{ds} = 2s + \frac{4\sqrt{3}}{9}(2s-18) = 0$

$s = \frac{8\sqrt{3}}{2 + \frac{8\sqrt{3}}{9}} = \frac{36\sqrt{3}}{9 + 4\sqrt{3}}$

Substituting in ( 1 ) and simplifying, we get

$a = \frac{108}{9 + 4\sqrt{3}}$

So size of two pieces are $4s$ and $3a$ with $s$ and $a$ with values as above.