# 28 - Solved problem in maxima and minima

**Problem 28**

The perimeter of an isosceles triangle is P inches. Find the maximum area.

**Solution:**

$P = 2y + x$

$0 = 2y' + 1$

$y' = -1/2 = -0.5$

Area:

$A = \frac{1}{2} xh$

$h = \sqrt{y^2 \, - \, 0.25x^2}$

Thus,

$A = \frac{1}{2}x \sqrt{y^2 - 0.25x^2}$

$\dfrac{dA}{dx} = \dfrac{1}{2} \left[ x \left( \dfrac{2y \, y' - 0.5x}{2 \sqrt{y^2 - 0.25x^2}} + \sqrt{y^2 - 0.25x^2}\right) \right] = 0$

$\dfrac{2(xy\,y' - 0.25x^2)}{2\sqrt{y^2 - 0.25x^2}} + \sqrt{y^2 - 0.25x^2} = 0$

multiply both sides of the equation by $\sqrt{y^2 - 0.25x^2}$

$xy\,y' - 0.25x^2 + y^2 - 0.25x^2 = 0$

$xy(-0.5) - 0.5x^2 + y^2 = 0$

$y^2 - 0.5xy - 0.5x^2 = 0$

$2y^2 - xy - x^2 = 0$

Solving for y by quadratic formula: a = 2; b = -x; c = -x^{2}

$y = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$y = \dfrac{-(-x) \pm \sqrt{x^2 - 4(2)(-x^2)}}{2(2)}$

$y = \dfrac{x \pm \sqrt{x^2 + 8x^2}}{4}$

$y = \dfrac{x \pm 3x}{4}$

$y = x \, \text{ and } \, -\frac{1}{2}x$

y = -½ x is absurd, thus use y = x

Therefore

$P = 2x + x = 3x$

$x = y = \frac{1}{3}P$

$h = \sqrt{y^2 \, - \, 0.25x^2}$

$h = \sqrt{(\frac{1}{3}P)^2 - \frac{1}{4}(\frac{1}{3}P)^2}$

$h = \sqrt{\frac{1}{9}P^2 - \frac{1}{36}P^2}$

$h = \sqrt{\frac{1}{12}P^2}$

$h = \frac{1}{2\sqrt{3}}P$

$A = \frac{1}{2}xh$

$A_{max} = \frac{1}{2}(\frac{1}{3}P)(\frac{1}{2\sqrt{3}}P)$

$A_{max} = \frac{1}{12\sqrt{3}}P^2$

$A_{max} = \frac{1}{12\sqrt{3}} \, \cdot \, \frac{\sqrt{3}}{\sqrt{3}}P^2$

$A_{max} = \frac{\sqrt{3}}{36}P^2$ *answer*