Two vertices of a rectangle are on the positive x-axis. The other two vertices are on the lines y = 4x and y = 6 - 5x. What is the maximum possible area of the rectangle?

For largest rectangle
$x = \frac{1}{2}(1.2) = 0.6$

$y = \frac{1}{2}(8/3) = 4/3$

 

$A_{max} = 0.6(4/3) = 0.8 ~ \text{unit}^2$

By ratio and proportion
$\dfrac{\frac{8}{3} - y}{x} = \dfrac{\frac{8}{3}}{1.2}$

$\frac{8}{3} - y = \frac{20}{9}x$

$y = \frac{8}{3} - \frac{20}{9}x$
 

 

Area of the rectangle
$A = xy$

$A = x \left( \frac{8}{3} - \frac{20}{9}x \right)$

$A = \frac{8}{3}x - \frac{20}{9}x^2$   ←   Eq. (1)
 

For maximum possible area of the rectangle:
$\dfrac{dA}{dx} = \frac{8}{3} - \frac{40}{9}x = 0$

$\frac{8}{3} = \frac{40}{9}x$

$x = \frac{3}{5} ~ \text{units}$
 

From Eq. (1)
$A_{max} = \frac{8}{3}\left( \frac{3}{5} \right) - \frac{20}{9}\left( \frac{3}{5} \right)^2$

$A_{max} = 0.8 ~ \text{unit}^2$