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

• dontdisturbmycircles
In summary, the author is trying to find the dimensions of an isoceles triangle which covers the most area inscribed in a circle of radius 4cm. They start by setting the bottom of the triangle to x, the height of the triangle to h, and drawing the radius of the circle, 4. They then 1/2 the base of the triangle to x, the height of the triangle at the bottom of the isoceles triangle, and get to a point where they have gotten their derivative of the area function. They simplify the expression for A to find that it equals 0 at h=0 and h=8. They are then able to multiply both sides by 2(8h-h^{2})^{1/2
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?

Last edited:
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?

Last edited:
Ah okay, that makes sense.

What do you mean I assumed it was not 8?

(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.

Last edited:
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.

## 1. How do you find the maximum area of an isosceles triangle inscribed in a circle with a radius of 4cm?

To find the maximum area of an isosceles triangle inscribed in a circle with a radius of 4cm, we can use the formula A = (1/2)r^2sin(2θ), where r is the radius of the circle and θ is the angle between the two equal sides of the triangle. We can then find the value of θ that maximizes the area by taking the derivative of the formula and setting it equal to 0. This will give us the value of θ that produces the maximum area.

## 2. What is an isosceles triangle?

An isosceles triangle is a triangle that has two equal sides and two equal angles. The base angles of an isosceles triangle are always equal, and the third angle is called the vertex angle. In an isosceles triangle, the vertex angle is always opposite the equal sides.

## 3. How do you inscribe a triangle in a circle?

To inscribe a triangle in a circle, we can draw a circle with a compass and then draw a triangle inside the circle, making sure that all three vertices of the triangle touch the circumference of the circle. This means that the sides of the triangle will be tangents to the circle.

## 4. What is the radius of a circle?

The radius of a circle is the distance from the center of the circle to any point on the circumference. It is represented by the letter "r" and is half the diameter of the circle.

## 5. What is the maximum area of an isosceles triangle inscribed in a circle with a radius of 4cm?

The maximum area of an isosceles triangle inscribed in a circle with a radius of 4cm is approximately 8.944cm^2. This can be found by using the formula A = (1/2)r^2sin(2θ) and finding the value of θ that maximizes the area, which is approximately 101.25 degrees. Plugging this value into the formula will give us the maximum area of the triangle.

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