I\'m having a tough time coming up with this one. I started with a circle on the

Question

I'm having a tough time coming up with this one. I started with a circle on the coordinate plane, with a center of (0,0) and radius r, then inscribed the triangle, with the top vertex (0,r). I made the vertexes at the base of the triangle (-x,-y) and (x,-y), so my base of the triangle is 2x. Then I used the height of the triangle as r+y.

From here I have the equations A=1/2bh or A=1/2(2x)(r+y) which equals x(r+y).
I also got the value of y as y=sq(r^2-x^2) because r^2=x^2+y^2.
From there, I plugged it back into the area function, and got A=x(r+sq(r^2-x^2))

When I set the area derivative = 0, I think I did something wrong, because I wasn't able to solve it. I got A'=r+(r^2-x^2)^(1/2)+x[1+1/2((r^2-x^2)^(-1/2))(-2x)], and I wasn't able to solve for A'=0.

Could someone walk me through the problem, I think I'm on the wrong track. Thanks

Explanation / Answer

The error is when you take derivative A'=r+(r^2-x^2)^(1/2)+x[1+1/2((r^2-x^2)^(-1/2))(-2x)] I think you used the product rule When you take the derivative of r+sq(r^2-x^2) with respect to x, r'=0, not 1 So you should have A'=r+(r^2-x^2)^(1/2)+x[1/2((r^2-x^2)^(-1/2))(-2x)] The rest is correct Redo it!

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