- #1

dontdisturbmycircles

- 592

- 3

Hi,

I am trying to use derivatives to find the dimensions of the isoceles triangle which covers the most area inscribed in a circle of radius 4cm.

So I set the 1/2 the bottom of the triangle to x, the height of the triangle to h, and draw the radius of the circle, 4, connecting the center of the circle and the endpoint of one side of the base.

1/2 the base = x

height of isoceles triangle = h

height of triangle drawn at bottom of isoceles triangle = y = h-4

So I get to a point where I have gotten my derivative of the area function, and am trying to find a point where it = 0 so that I can find out whether that point is a max and solve the problem.

I get [tex]\frac{dA}{dh}=h*\frac{1}{2}(8h-h^{2})^{-1/2}(8-2h)+(8h-h^{2})^{1/2}=0[/tex]

Then I simplify it to [tex]\frac{8h-2h^{2}+2(8h-h^{2})}{2(8h-h^{2})^{1/2}}=0[/tex]

At this point can I multiply both sides by [tex]2(8h-h^{2})^{1/2}[/tex] to get

[tex]8h-2h^{2}+16h-2h^{2}=0[/tex]?

I am usually hesitant to multiply by a variable when there is a 0 sitting on the other side, but I am thinking that since h must be > 0, I can do it. Am I wrong in this assumption?

I guess my broader question is when is it okay to multiply by a variable when there is a 0 on the other side?

I am trying to use derivatives to find the dimensions of the isoceles triangle which covers the most area inscribed in a circle of radius 4cm.

So I set the 1/2 the bottom of the triangle to x, the height of the triangle to h, and draw the radius of the circle, 4, connecting the center of the circle and the endpoint of one side of the base.

1/2 the base = x

height of isoceles triangle = h

height of triangle drawn at bottom of isoceles triangle = y = h-4

So I get to a point where I have gotten my derivative of the area function, and am trying to find a point where it = 0 so that I can find out whether that point is a max and solve the problem.

I get [tex]\frac{dA}{dh}=h*\frac{1}{2}(8h-h^{2})^{-1/2}(8-2h)+(8h-h^{2})^{1/2}=0[/tex]

Then I simplify it to [tex]\frac{8h-2h^{2}+2(8h-h^{2})}{2(8h-h^{2})^{1/2}}=0[/tex]

At this point can I multiply both sides by [tex]2(8h-h^{2})^{1/2}[/tex] to get

[tex]8h-2h^{2}+16h-2h^{2}=0[/tex]?

I am usually hesitant to multiply by a variable when there is a 0 sitting on the other side, but I am thinking that since h must be > 0, I can do it. Am I wrong in this assumption?

I guess my broader question is when is it okay to multiply by a variable when there is a 0 on the other side?

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