Find the area of the largest rectangle that can be inscribed in a right triangle with legs of length 3 cm and 4 cm if two sides of the rectangle lie along the legs.

Area of largest rectangle:
$A_{max} = \frac{1}{2}A_{tri}$
 

 

$A_{max} = \frac{1}{2} \times \frac{1}{2}(4)(3)$

$A_{max} = 3 ~ \text{cm}^2$

Area of the rectangle :
$A = xy$
 

 

From the figure above:
$\dfrac{y}{4 - x} = \dfrac{3}{4}$

$y = \frac{3}{4}(4 - x)$
 

Hernce,
$A = x \cdot \frac{3}{4}(4 - x)$

$A = \frac{3}{4}(4x - x^2)$
 

For largest rectangle:
$\dfrac{dA}{dx} = \frac{3}{4}(4 - 2x) = 0$

$x = 2$
 

Largest rectangle:
$A_{max} = \frac{3}{4}\left[ 4(2) - 2^2 \right]$

$A_{max} = 3 ~ \text{cm}^2$