1. ## help please optimization!

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

Help!

2. 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 )

3. Ok but any tips on how to do it with calculus?

4. 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.

5. 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.

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