32 - 34 Maxima and minima problems of a rectangle inscribed in a triangle

From the figure:
$\dfrac{y}{a - x} = \dfrac{b}{a}$

$y = \dfrac{b}{a}(a - x)$
 

$A = x \dfrac{b}{a}(a - x)$

$A = bx - \dfrac{b}{a}x^2$

$\dfrac{dA}{dx} = b - \dfrac{2b}{a}x = 0$

$\dfrac{2b}{a}x = b$

$x = \frac{1}{2}a$
 

$y = \dfrac{b}{a}(a - \frac{1}{2}a)$

$y = \frac{1}{2}b$
 

Dimensions: ½ a × ½ b          answer

 
By similar triangle:
$\cot \alpha = \dfrac{r}{x} = \dfrac{60}{80}$

$r = \frac{3}{4}x$
 

$\cot \beta = \dfrac{s}{x} = \dfrac{80}{60}$

$s = \frac{4}{3}x$
 

$r + y + s = 100$

$\frac{3}{4}x + y + \frac{4}{3}x = 100$

$y = 100 - \frac{25}{12}x$
 

Thus,
$A = x(100 - \frac{25}{12}x)$

$A = 100x - \frac{25}{12}x^2$

$\dfrac{dA}{dx} = 100 - \frac{50}{12}x$

$x = 100(\frac{12}{50}$

$x = 24 \, \text{ feet }$
 

$y = 100 - \frac{25}{12}(24)$

$y = 50 \, \text{ feet }$
 

Dimensions: 50 ft × 24 ft           answer

By similar triangle:
$\cot \alpha = \dfrac{r}{x} = \dfrac{b}{a}$

$r = \dfrac{b}{a}x$
 

$\cot \beta = \dfrac{s}{x} = \dfrac{a}{b}$

$s = \dfrac{a}{b}x$
 

$r + y + s = \sqrt{a^2 + b^2}$

$\frac{b}{a}x + y + \dfrac{a}{b}x = \sqrt{a^2 + b^2}$

$y + \dfrac{a^2 + b^2}{ab}x = \sqrt{a^2 + b^2}$

$y = \sqrt{a^2 + b^2} - \dfrac{a^2 + b^2}{ab}x$

$y = \dfrac{ab\sqrt{a^2 + b^2} - (a^2 + b^2)x}{ab}$

$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - x \sqrt{a^2 + b^2} \right)$
 

Thus,
$A = x \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - x \sqrt{a^2 + b^2} \right)$

$A = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( abx - x^2 \sqrt{a^2 + b^2} \right)$

$\dfrac{dA}{dx} = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - 2x \sqrt{a^2 + b^2} \right) = 0$

$2x \sqrt{a^2 + b^2} = ab$

$x = \dfrac{ab}{2 \sqrt{a^2 + b^2}}$
 

$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - \dfrac{ab}{2 \sqrt{a^2 + b^2}} \cdot \sqrt{a^2 + b^2} \right)$

$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( ab - \dfrac{ab}{2} \right)$

$y = \dfrac{\sqrt{a^2 + b^2}}{ab} \left( \dfrac{ab}{2} \right)$

$y = \dfrac{\sqrt{a^2 + b^2}}{2}$
 

Dimensions:
$\dfrac{ab}{2 \sqrt{a^2 + b^2}} \, \times \, \dfrac{\sqrt{a^2 + b^2}}{2}$           answer