Optimization parabola/triangle

mattc · #1 $Old$ April 26th, 2009, 07:29 PM

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?

skeeter · #2 $Old$ April 26th, 2009, 07:34 PM

Quote:

Originally Posted by mattc $View Post$

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?

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

$A(x) = \frac{1}{2}(2x)(y) = x(4-x^2)$

mattc · #3 $Old$ April 26th, 2009, 08:07 PM

great thanks I was afraid it was that simple.

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

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

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)?

mattc · #4 $Old$ April 26th, 2009, 09:11 PM

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?