• Dec 19th 2009, 11:57 AM
amma0913
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 PM
drop10
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 PM
amma0913
Ok but any tips on how to do it with calculus?
• Dec 19th 2009, 12:30 PM
drop10
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 PM
mr fantastic
Quote:

Originally Posted by amma0913
Ok but any tips on how to do it with calculus?

You were given the rule for area as a function of x - the hard part of the question. If you're attempting questions like this then you should already know how to use calculus to find the maximum of a function.