A corner lot, triangular in shape, has perpendicular sides measuring 90 m and 120 m, respectively.

Maximum area of rectangle
$A_{max} = \frac{1}{2}A_{tri}$
 

 

$A_{max} = \frac{1}{2} \left[ \frac{1}{2}(120)(90) \right]$

$A_{max} = 2700 ~ \text{m}^2$

$\cot \theta$:

$\dfrac{a}{y} = \dfrac{120}{90}$

$a = \frac{4}{3}y$

 

$\tan \theta$:

$\dfrac{b}{y} = \dfrac{90}{120}$

$b = \frac{3}{4}y$

 

$x = 150 - \frac{4}{3}y - \frac{3}{4}y$

$x = 150 - \frac{25}{12}y$
 

 

Area of rectangle:
$A = xy$

$A = \left( 150 - \frac{25}{12}y \right) y$

$A = 150y - \frac{25}{12}y^2$
 

For largest area of the building
$\dfrac{dA}{dy} = 150 - \frac{25}{6}y = 0$

$150 = \frac{25}{6}y$

$y = 36 ~ \text{m}$
 

Hence,
$A_{max} = 150(36) - \frac{25}{12}(36^2)$

$A_{max} = 2700 ~ \text{m}^2$