# Optimization parabola/triangle

• Apr 26th 2009, 05:29 PM
mattc
Optimization parabola/triangle
An isoceles triangle with vertices (0,0), (x,y) & (-x,y) is inscribed in the parabola y=4-x^2

What x & y will give triangle max area and what is that area?

I'm unsure of my first move to get an A(x).

Edit: also to I need to split my triangle into halves for this?
• Apr 26th 2009, 05:34 PM
skeeter
Quote:

Originally Posted by mattc
An isoceles triangle with vertices (0,0), (x,y) & (-x,y) is inscribed in the parabola y=4-x^2

What x & y will give triangle max area and what is that area?

I'm unsure of my first move to get an A(x).

Edit: also to I need to split my triangle into halves for this?

$\displaystyle A = \frac{1}{2}bh$

$\displaystyle A(x) = \frac{1}{2}(2x)(y) = x(4-x^2)$
• Apr 26th 2009, 06:07 PM
mattc
great thanks I was afraid it was that simple.

I found the A'(x) to get to the CV points.

$\displaystyle A'(x) = -3x^2+4$

I know important parts of the parabola are max at (0,4) and y=0 at x=-2 & 2. So I know where the triangle contained. Do I plug the x values into A'(x)?
• Apr 26th 2009, 07:11 PM
mattc
I ended up with (-1.15, 2.68) and (1.15, 2.68) and an area of 3.08 units cubed. My graph of A'(x) and critical points with the original function all seem to fit. Is this correct?