# 60 - 61 Maxima and minima problems of a folded page

**Problem 60**

One corner of a leaf of width a is folded over so as just to reach the opposite side of the page. Find the width of the part folded over when the length of the crease is a minimum. See Figure 41.

**Solution:**

$\alpha = 90^\circ - \theta$

$\beta = 180^\circ - 2\theta$

$\beta = 2(90^\circ - \theta)$

$\beta = 2\alpha$

$\cos \beta = \dfrac{a - x}{x}$

$\cos 2\alpha = \dfrac{a - x}{x}$

$\cos^2 \alpha - \sin^2 \alpha = \dfrac{a - x}{x}$ ← from double angle formula

$\left( \dfrac{\sqrt{c^2 - x^2}}{c} \right)^2 - \left( \dfrac{x}{c} \right)^2 = \dfrac{a - x}{x}$

$\dfrac{c^2 - x^2}{c^2} - \dfrac{x^2}{c^2} = \dfrac{a - x}{x}$

$\dfrac{c^2 - 2x^2}{c^2} = \dfrac{a - x}{x}$

$c^2 \, x - 2x^3 = ac^2 - c^2 \, x$

$(2x - a)c^2 = 2x^3$

$c = \sqrt{\dfrac{2x^3}{2x - a}}$

$\dfrac{dc}{dx} = \dfrac{\dfrac{(2x - a)(6x^2) - 2x^3 (2)}{(2x - a)^2}}{2\sqrt{\dfrac{2x^3}{2x - a}}} = 0$

$3(2x - a) - 2x = 0$

$4x = 3a$

$x = \frac{3}{4}a$ *answer*

**Problem 61**

Solve Problem 60 above if the area folded over is to be a minimum.

**Solution:**

$c = \sqrt{\dfrac{2x^3}{2x - a}}$

$c^2 = \dfrac{2x^3}{2x - a}$

Thus,

$\sqrt{c^2 - x^2} = \sqrt{\dfrac{2x^3}{2x - a} - x^2}$

$\sqrt{c^2 - x^2} = \sqrt{\dfrac{2x^3 - x^2 (2x - a)}{2x - a}}$

$\sqrt{c^2 - x^2} = x \, \sqrt{\dfrac{a}{2x - a}}$

Area:

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

$A = \frac{1}{2}x^2 \sqrt{\dfrac{a}{2x - a}}$

$\dfrac{dA}{dx} = \frac{1}{2} \left[ x^2 \dfrac{\dfrac{-a(2)}{(2x - a)^2}}{2\sqrt{\dfrac{a}{2x - a}}} + 2x \sqrt{\dfrac{a}{2x - a}} \right] = 0$

$2 \sqrt{\dfrac{a}{2x - a}} = \dfrac{ax}{(2x - a)^2 \sqrt{\dfrac{a}{2x - a}}}$

$\dfrac{2a}{2x - a} = \dfrac{ax}{(2x - a)^2}$

$2(2x - a) = x$

$3x = 2a$

$x = \frac{2}{3}a$ *answer*