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## Homework Statement

Consider the part of the parabola y=1-x^2 from x=-1 to x=1. This curve fits snugly inside an isosceles triangle with base on the x-axis and one vertex on the y-axis. What is the smallest possible area of such a triangle?

## The Attempt at a Solution

[tex] A = \frac{1}{2}x y[/tex]

[tex] y = 1-x^2 [/tex]

so

[tex] A = \frac{1}{2}x(1-x^2)[/tex]

I solved that, took the derivative and set it equal to 0. That gave me an answer of x=sqrt(1/3), but I know that is wrong since it doesn't encompass the parabola curve. What is wrong with this?