# Homework Help: Optimization of the Area of a Triangle

1. Oct 6, 2014

### reigner617

1. The problem statement, all variables and given/known data
Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 4. Solve by writing the area as a function of θ

2. Relevant equations
A=1/2 (bh)

3. The attempt at a solution

Given the side h and the hypotenuse 4, we can find the base of the smallest triangle to be sqrt(16-h2) which means that the base of the largest triangle is 2(sqrt(16-h2)). Its height can be considered as h+4, which would give a side length of sqrt(8h+32). So the area would be:
sqrt(16-h2)(h+4). However, they want it to be solved in terms of theta, which is the part that I am not sure of.

2. Oct 6, 2014

### BiGyElLoWhAt

can you express h in terms of theta? express the sides of the triangle in terms of theta.

3. Oct 6, 2014

### reigner617

for convenience, I'll say c= sqrt(8h+32). Would it be correct if I said the height was c(sinθ) and the base of the largest triangle 2c(cosθ)?

4. Oct 6, 2014

### LCKurtz

I don't know where you got $c = \sqrt{8h+32}$. It would help if you labeled your picture. If you note that the angle between the two sides that you have labeled $4$ and $h$ is $2\theta$ that will help you get the two legs of that little right triangle in terms of $\theta$.

5. Oct 6, 2014

### reigner617

How did you determine that angle to be 2θ?
And to answer your question, I used the pythagorean theorem to find one of the side lengths

6. Oct 6, 2014

### RUber

From what you have above, it seems that the height $(h+4)= Ccos(\theta)$ Where, as you mentioned, C is the length of the hypotenuse. You should be able to solve that for h in terms of theta and plug that into the formula for area.

7. Oct 6, 2014

### reigner617

If I solve for h, that would give me Ccosθ-4=h. Is it ok to have an extra variable (C) in there?

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8. Oct 6, 2014

### LCKurtz

That upper triangle is an isosceles triangle with equal sides of $4$. Its angles are $\theta,\theta, 180 -2\theta$. Label them. The supplement of that last one is the angle you are looking for.

Once you have that you have both $h$ and the other leg in terms of $\theta$.

9. Oct 6, 2014

### reigner617

Are these labeled correctly?

If those are correct, can I use the law of sines such that

h/(sin90-2θ) = 4/sin90
h = 4(sin90-2θ)
h=4(1-2θ)
h=4-8θ

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10. Oct 6, 2014

### LCKurtz

You mean $h/\sin(90-2\theta)$. Parentheses are important.

Which is why you then get nonsense. Also, it is silly to use the law of sines given the picture. You have adjacent sides to the $h$ and $4$ for the angle $2\theta$. What trig function does that give you for $h$? Similarly for the other leg.

11. Oct 6, 2014

### reigner617

That would be cos2θ=h/4
4cos2θ=h

12. Oct 6, 2014

### LCKurtz

Yes. And what about the other leg?

13. Oct 6, 2014

### reigner617

Well since h=4cos2θ, couldn't we just substitute it into the expression to get sqrt(16-(4cos2θ)^2) for the base?

14. Oct 6, 2014

### RUber

That looks a lot like $4(\sqrt(1-\cos^2 2\theta)$. It seems like the long way to get to the opposite side from the angle $2\theta$ with hypotenuse 4.

15. Oct 6, 2014

### LCKurtz

Yes, you could, but you would want to simplify it. But, again, it is silly to do it that way. What trig formula uses opposite/hypotenuse?

16. Oct 6, 2014

### reigner617

That would be sine. So if I were to use that, I would get sin(2θ)= [sqrt(16-(h^2))]/4. But I don't understand what this would accomplish since we're trying to look for max area, and that involves base and height

17. Oct 6, 2014

### reigner617

Ahhh never mind. I think I see it now

18. Oct 6, 2014

### LCKurtz

So instead of calling that leg $\sqrt{16-h^2}$ you can call it $4\sin(2\theta)$. Now you have both legs of that little triangle in terms of $\theta$. Now you can express the area of the original big triangle in terms of $\theta$. Then you are ready for the calculus problem of maximizing the area. Too bad the trig is holding you back so badly. Perhaps you should review it.

19. Oct 6, 2014

### reigner617

So A= 1/2 (4cos(2θ))(4sin(2θ))
= 8cos(2θ)sin(2θ)

deriving this, we get A'= -16(sin(2θ))2 + 16(cos(2θ))2

setting it to zero, we get θ= 30 degrees

substituting this for θ we get 2(sqrt(3))

However, another part of the question was solving the area in terms of h (I used A= [sqrt(16-h2)](h+4)

Following the same steps as I did for the one above, I get h=-4 and 2. h can't be -4 because if we use it, h+4 would go to zero, and the whole area would be zero. So h can only be 2.

Substituting this for h I get 12(sqrt(3))

What am I doing wrong?

20. Oct 6, 2014

### LCKurtz

Isn't the height of your triangle $4+4\cos(2\theta)$? And why the 1/2 out in front?

21. Oct 6, 2014

### reigner617

Ah thank you for pointing that out. It works now. The 1/2 was put there by mistake because I was thinking fo 1/2 (bh). Thank you for all your help

22. Oct 6, 2014

### LCKurtz

It works now? Given what I have seen so far, I doubt it. How about showing what you did to finish?

23. Oct 6, 2014

### reigner617

I used 4cos(2x)+4 for the height so A=4sin2x(4+4cos2x). I derived that and still got 30 for the angle. Substituting that into the Area formula, I get 12(sqrt(3)) which is the same as the answer I got when I solved in terms of h

24. Oct 7, 2014

### RUber

$4\sin 2\theta$ is 1/2 base and $4 + 4 \cos 2\theta$ is the height.
Your method of taking the derivative and setting it to zero is good. The best option is to rearrange the derivative as a quadratic function of $\cos 2 \theta$. h=-4 will be your minimum (zero) and another value for h will be your max. Both should be solutions of A'=0.

25. Oct 7, 2014

### LCKurtz

Good job. I presume you have noticed that the optimal shape is an equilateral triangle.