1. Hey Everyone,

I'm having trouble setting up this word problem.
I know that area of a circle is pir2 and area of a triangle is 1/2bh but for some reason I cant find a way to combine these equation together.

What are the radius and area of the circle of maximum area that can be inscribed in an isosceles triangle whose two equal sides have length 1?

2.

3. This is how I approached the problem:

Bisect the isosceles triangle into two right triangles, and put one of the right triangles in the $xy$-plane with the right angle at the origin, as in the diagram:

The vertical leg has length $b$ and the horizontal leg has length $b$.

The hypotenuse of the triangle lies along the line $y(x)$. By the two-intercept form of a line, we know:

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$

By Pythagoras, and the fact that the hypotenuse is 1 unit in length, we know:

$\displaystyle a^2+b^2=1$

This will allow you to write the line in terms of one parameter.

Next, use the fact that the perpendicular distance from the point $(r,0)$ to this line is simply $r$. This will allow you to write $r$ as a function of one parameter. If you express the line in slope-intercept form, you may use the formula for the perpendicular distance between a point and a line:

$\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}$

Once you have $r$ as a function of one parameter, differentiate $r$ with respect to this parameter, and equate to zero to find the critical value within the domain. Then evaluate $r$ at this critical value to find the maximum radius, and plug this into the formula for the area of the circle having this radius.

Please feel free to post your work if you get stuck or need further guidance.

4. Originally Posted by dexstarr
Hey Everyone,

I'm having trouble setting up this word problem.
I know that area of a circle is pir2 and area of a triangle is 1/2bh but for some reason I cant find a way to combine these equation together.

What are the radius and area of the circle of maximum area that can be inscribed in an isosceles triangle whose two equal sides have length 1?
I'd advise you to start by drawing out the situation. Call the third length of your triangle \displaystyle \displaystyle \begin{align*} x \end{align*}. Can you find a way to write the radius of your circle in terms of \displaystyle \displaystyle \begin{align*} x \end{align*}?

Also, it might be more useful to use Heron's Formula for the area of a triangle: If your sides are \displaystyle \displaystyle \begin{align*} a,b,c \end{align*}, then the semiperimeter is \displaystyle \displaystyle \begin{align*} s = \frac{a + b + c}{2} \end{align*} and the area of your triangle is \displaystyle \displaystyle \begin{align*} A = \sqrt{ s ( s - a)( s- b) (s - c)} \end{align*}.

s = (2 + x)/(2) or (1 + x/2)

r^2 = [((1 + x/2)-1)((1 + x/2)-1)((1 + x/2)-x)] / (1 + x/2)

r^2 = (2x^2 - x^3) / (4x+8)

r^2 ' = (-x^3-2x^2+4x) / (2x^2+8x+8)

(-x^3-2x^2+4x) / (2x^2+8x+8) = 0

Max is x= -1 + (5)^1/2
Min is x= 0

Is this correct so far?

6. Your result agrees with the work I did, as your value for $x$ is twice what I found for $b$. I will admit though, that I do not follow what you did to get there.

7. Hi,
can you post the final answer to this problem so that I can check it with mine?

Originally Posted by MarkFL
This is how I approached the problem:

Bisect the isosceles triangle into two right triangles, and put one of the right triangles in the $xy$-plane with the right angle at the origin, as in the diagram:

The vertical leg has length $b$ and the horizontal leg has length $b$.

The hypotenuse of the triangle lies along the line $y(x)$. By the two-intercept form of a line, we know:

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$

By Pythagoras, and the fact that the hypotenuse is 1 unit in length, we know:

$\displaystyle a^2+b^2=1$

This will allow you to write the line in terms of one parameter.

Next, use the fact that the perpendicular distance from the point $(r,0)$ to this line is simply $r$. This will allow you to write $r$ as a function of one parameter. If you express the line in slope-intercept form, you may use the formula for the perpendicular distance between a point and a line:

$\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}$

Once you have $r$ as a function of one parameter, differentiate $r$ with respect to this parameter, and equate to zero to find the critical value within the domain. Then evaluate $r$ at this critical value to find the maximum radius, and plug this into the formula for the area of the circle having this radius.

Please feel free to post your work if you get stuck or need further guidance.

8. Originally Posted by dmireya
Hi,
can you post the final answer to this problem so that I can check it with mine?
I no longer have my notes on this problem, but if you want to post your work, I will be glad to go over it.

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