Find the maximal area of a rectangle inscribed in a right triangle whose two lengs have length 6 and 8

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- Dec 19th 2009, 11:57 AMamma0913help please optimization!
Find the maximal area of a rectangle inscribed in a right triangle whose two lengs have length 6 and 8

Help! - Dec 19th 2009, 12:09 PMdrop10
**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 (Wink)) - Dec 19th 2009, 12:12 PMamma0913
Ok but any tips on how to do it with calculus?

- Dec 19th 2009, 12:30 PMdrop10
Yea, sure!

Instead of finding the vertex using $\displaystyle \frac{-b}{2a}$,take the derivative and find where it is zero and you will get the same answer. - Dec 19th 2009, 02:10 PMmr fantastic