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

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- Apr 22nd 2008, 03:37 AMsomeone21Optimization
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 - Apr 22nd 2008, 06:41 AMmr fantastic
- Apr 22nd 2008, 07:37 AMsomeone21
- Apr 22nd 2008, 03:35 PMmr fantastic
- Apr 22nd 2008, 03:50 PMmr 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:

$\displaystyle K = \frac{a^2 b}{8}$ .... (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. - Apr 22nd 2008, 04:16 PMmr 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 $\displaystyle \phi$.

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

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 $\displaystyle \frac{\phi}{2}$ and the perpendicular is the height of the triangle.

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

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

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

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

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

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

Therefore:

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

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

Note: I reserve the right for this reply to contain algebraic errors. - Apr 23rd 2008, 01:58 AMsomeone21
- Apr 23rd 2008, 01:59 AMsomeone21
- Apr 23rd 2008, 04:25 AMmr fantastic
- Apr 23rd 2008, 05:08 AMmr fantastic
Let 2

*b*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:

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

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

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

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 $\displaystyle 2b = 4 \sqrt{3} = a$ and so the triangle is equilateral.

Note: I reserve the right for this reply to contain careless errors. - Apr 23rd 2008, 05:26 AMcolby2152
**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.

$\displaystyle \lim_{s \rightarrow 0} A$

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

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

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. - Apr 23rd 2008, 07:06 AMsomeone21
- May 5th 2008, 11:10 AMcolby2152