Find the maximal area of a rectangle inscribed in a right triangle whose two lengs have length 6 and 8
Help!
Given: Need to find the maximum area of a rectangle inscribed in a right-triangle with legs with lengths 6 and 8 respectively.
Solution:
Find an equation for the boundary lines:
$\displaystyle
x=0
$
$\displaystyle
y=0
$
$\displaystyle
y=8-\frac{4}{3}x
$
Now, find the equation of the area of the circle:
$\displaystyle
Area(x,y) = x*y
$
$\displaystyle
Area(x) = x*(8-\frac{4}{3}x)
$
$\displaystyle
Area(x) = 8x - \frac{4}{3}x^2
$
Now find the vertex $\displaystyle \frac{-b}{2a} = \frac{-8}{2*(\frac{-4}{3})} = \frac{4}{\frac{4}{3}} = 3$
Substitute $\displaystyle x=3$ into the equation for area to get your answer (I'll let you do it yourself )