Thread: 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?

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?

great thanks I was afraid it was that simple.

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



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)?

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?