Optimization

**someone21** · April 22nd 2008, 03:37 AM

An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well

**mr fantastic** · April 22nd 2008, 06:41 AM

Originally Posted by someone21

An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well

Minimum or maximum? The minimum area is trivially zero.

**someone21** · April 22nd 2008, 07:37 AM

Originally Posted by mr fantastic

Minimum or maximum? The minimum area is trivially zero.

Minimum possible area but why is it zero??

And if maximum possible area then how to do it???

**mr fantastic** · April 22nd 2008, 03:35 PM

Originally Posted by someone21

Minimum possible area but why is it zero??

And if maximum possible area then how to do it???

Draw an isosceles triangle with non-equal side the diameter of the circle. Now move the non-equal side up so that the two equal sides get smaller and smaller. The triangle degenerates to a point - of area zero.

**mr fantastic** · April 22nd 2008, 03:50 PM

Originally Posted by someone21

An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well

Scroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles

Your circle has a radius r = 2. Let a = c and re-arrange:

$K = \frac{a^2 b}{8}$ .... (1)

where K is the area of the triangle.

You need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....

I haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.

**mr fantastic** · April 22nd 2008, 04:16 PM

Originally Posted by mr fantastic

Scroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles

Your circle has a radius r = 2. Let a = c and re-arrange:

$K = \frac{a^2 b}{8}$ .... (1)

where K is the area of the triangle.

You need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....

I haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.

Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:

Let the angle at the top vertex of the triangle be of size $\phi$ .

Let the two equal angles at the base of the triangle be of size $\theta$ .

Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.

Area of any triangle: $A = \frac{bh}{2}$ .

To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).

You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$ ). Then:

$a = 4 \cos(\phi/2)$ .

$b = 8 \cos(\phi/2) \sin(\phi/2)$ .

$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$ .

Therefore:

$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$ .

Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$ , that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.

Note: I reserve the right for this reply to contain algebraic errors.

**someone21** · April 23rd 2008, 01:58 AM

Originally Posted by mr fantastic

Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:

Let the angle at the top vertex of the triangle be of size $\phi$ .

Let the two equal angles at the base of the triangle be of size $\theta$ .

Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.

Area of any triangle: $A = \frac{bh}{2}$ .

To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).

You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$ ). Then:

$a = 4 \cos(\phi/2)$ .

$b = 8 \cos(\phi/2) \sin(\phi/2)$ .

$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$ .

Therefore:

$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$ .

Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$ , that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.

Note: I reserve the right for this reply to contain algebraic errors.

I think the question was the circle is inside the isoceles triangle?

**someone21** · April 23rd 2008, 01:59 AM

Originally Posted by mr fantastic

Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:

Let the angle at the top vertex of the triangle be of size $\phi$ .

Let the two equal angles at the base of the triangle be of size $\theta$ .

Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.

Area of any triangle: $A = \frac{bh}{2}$ .

To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).

You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$ ). Then:

$a = 4 \cos(\phi/2)$ .

$b = 8 \cos(\phi/2) \sin(\phi/2)$ .

$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$ .

Therefore:

$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$ .

Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$ , that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.

Note: I reserve the right for this reply to contain algebraic errors.

I think the question was the circle is inside the isoceles triangle?

**mr fantastic** · April 23rd 2008, 04:25 AM

Originally Posted by someone21

I think the question was the circle is inside the isoceles triangle?

Ah yes. I misread circumscribed as inscribed. Oh well, I guess I answered the second part:

"Any other good notes or questions for Optimization problems would be welcomed as well"

**mr fantastic** · April 23rd 2008, 05:08 AM

Originally Posted by someone21

An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well

Let 2b be the length of the base of the triangle, a the length of the other two sides and h the height of triangle.

Then A = bh.

A relationship between b and h is need so that A can be expressed in terms of a single variable.

Draw a line from the middle of the base to the (top) vertex of the triangle. This line goes through the center of the circle and is the height of the triangle.

Draw a line from the center of the circle to one of the sides of the triangle. This line is perpendicular to that side (why?).

Apply Pythagoras Theorem to the two obvious right-triangles:

$h^2 + b^2 = a^2$ .... (1)

$2^2 + {\color{red}(a-b)}^2 = (h - 2)^2 \Rightarrow 4 + a^2 - 2ab + b^2 = h^2 - 4h + 4 \Rightarrow a^2 - 2ab + b^2 = h^2 - 4h$ .... (2)

You should consider the geometry that gives the red length .....

Solve equations (1) and (2) simultaneously for b in terms of h (a small bit of algebra for you to do):

$b = \frac{2h}{\sqrt{h^2 - 4h}}$ .

Therefore A = .......

Solve dA/dh = 0 (I get h = 6). Hence the minimum area of the triangle is equal to .....

Note: As you might anticipate, if you do the calculations you get $2b = 4 \sqrt{3} = a$ and so the triangle is equilateral.

Note: I reserve the right for this reply to contain careless errors.

**colby2152** · April 23rd 2008, 05:26 AM

Originally Posted by someone21

Minimum possible area but why is it zero??

One way to think about it....

An equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.

$\lim_{s \rightarrow 0} A$

$\lim_{s \rightarrow 0} \frac{1}{2\sqrt{2}}s^2 \rightarrow 0$

Since the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$

The limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.

**someone21** · April 23rd 2008, 07:06 AM

Originally Posted by colby2152

One way to think about it....

An equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.

$\lim_{s \rightarrow 0} A$

$\lim_{s \rightarrow 0} \frac{1}{2\sqrt{2}}s^2 \rightarrow 0$

Since the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$

The limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.

why is it s goes to zero and can it be explained more detaily please

**colby2152** · May 5th 2008, 11:10 AM

Originally Posted by someone21

why is it s goes to zero and can it be explained more detaily please

You are looking at the area as a side goes to zero. The area, dependent upon the sides, then goes to zero as well.

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