Solve Max Area Isoceles Triangle Inscribed in Circle of Radius 4cm

[dontdisturbmycircles]

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 \frac{dA}{dh}=h*\frac{1}{2}(8h-h^{2})^{-1/2}(8-2h)+(8h-h^{2})^{1/2}=0

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

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

8h-2h^{2}+16h-2h^{2}=0?

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?

 

[Hurkyl]

The only danger of multiplying is that you might be introducing extra solutions.

If you're worried, think back to what you know about rational numbers:

a/b = 0

if and only if

a = 0 and b \neq 0

(If b = 0, then you've done something wrong when deriving a/b=0)


Incidentally, somewhere along the line you assumed that h was neither 0 nor 8. Was that intentional?

 

[dontdisturbmycircles]

Ah okay, that makes sense.

What do you mean I assumed it was not 8?

 

[Hurkyl]

(foreward: your work is good, I don't mean to imply otherwise. I'm just bringing up some things to make you aware of them! So hopefully you won't be tricked in later problems where they matter)


In your expression for A, notice that you assumed that

0 <= h <= 8.

That's a good assumption to make, because your triangle is supposed to be inscribed in your circle. But, your expression for A is not differentiable at h=0 and h=8. So, the analysis you've presented is only valid for the range

0 < h < 8.


Remember that, to find the maximum of a continuous function over a closed interval, you need to consider three kinds of points:
(1) Points where the derivative is zero
(2) Points where the derivative does not exist
(3) Boundary points.

(remember that "critical point" includes points of type (1) and type (2))


So far, you've only considered points of type (1), and haven't looked at points of type (2) or type (3), so you're not quite done. Fortunately, those points are easy to handle.

 

[dontdisturbmycircles]

Ahhh that is very true. No worries I took no offense and didn't think you were being offensive, I was genuinely interested . I actually forgot that critical points include undefined derivatives so thanks alot. I appreciate it.