Question

A triangle has a base 12ft long and an altitude 8ft high. Find the area of the largest rectangle that can be inscribed in the triangle so that the base of the rectangle falls on the base of the triangle.
-ans: 24 sq. ft.

Answer 1

Hello,

Let's call x the length of the rectangle and y its height.

Using the theorem of Thales,

$\dfrac{8-y}{8} = \dfrac{x}{12} \\ ==\textgreater \ 8x=96-12y\\ ==\textgreater \ y= 8 -\dfrac{2}{3} x\\ Aera=A(x)=x*y=x*( 8 -\dfrac{2}{3} x)=\dfrac{24}{3}*x-\dfrac{2}{3}*x^2\\ =-\dfrac{2}{3}*(x^2-12x)\\ =-\dfrac{2}{3}*(x^2-12x+36-36)\\ =-\dfrac{2}{3}*(x-6)^2+24\\ =24-\dfrac{2}{3}*(x-6)^2\\ =\dfrac{2}{3}*(72-2(x-6)^2)$
$\mbox{which is max if x=6 then y=8 - \dfrac{2}{3} *6=4}$

Max Area=6*4=24 (ft²)