Optimize Smallest Area Triangle in Parabola y=1-x^2

[Sheneron]

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

A = \frac{1}{2}x y

y = 1-x^2

so

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

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?

 

[Firepanda]

y should be the point of the vertex not the equation of the parabola, try finding the gradient of the slopes at x=-1 and 1 and see where they cross.

 

[Sheneron]

Of which slopes, the parabolas?

 

[Dick]

Suppose the point of tangency is at x=a. Then you know the tangent line goes through the point (a,1-a^2) and has slope -2a, right? Work back from that and figure out where that tangent line intersects the x and y axes. Then find the area of the triangle as a function of a. Now minimize with respect to a. That's what Firepanda meant.

 

[Sheneron]

Ok here we go:

the slope is equal to rise over run so:
f'(a) = \frac{y-b}{x-a}
where a and b are just arbitrary points.

f'(p) = -2p = \frac{y - (1-p^2)}{x-p}

where p is some point. And f'(p) is the derivative of f at point p. We don't use x since x is already a unknown variable in our equation.

If we solve that for y we get:

y = -2p(x-p) + 1-p^2

y = p^2 -2px +1

This is our equation of some tangent line to the parabola at some point p.

Solve for the y intercept by setting y=0
0 = p^2 -2px +1

x = \frac{p^2 +1}{2p}

and we know the x intercept equals:

y = p^2 + 1

So back to our Area formula:

A(p) = \frac{1}{2}(\frac{p^2+1}{2p})(p^2+1)

Take the derivative of that and set it equal to 0.

Solve for p and you will get an answer of

p=\sqrt{\frac{1}{3}}

Plug that back into your x and y intercept equations to get the x and y values. Plug those x and y values back into the area equation to get an answer of about 1.5396.

 

[Dick]

Bravo! Well done! Except that I think you are forgetting the part of the triangle in the negative x region. Isn't the area of the triangle really x*y, not (1/2)*x*y? I.e. (1/2)*(2x)*y? Actually, I think you are right numerically. It's just that A(p) isn't what you said it is. Just a typo, I'm shure.