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Math Help - Maximum Area

  1. #1
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    Exclamation Maximum Area

    Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 4.

    I know how to maximize, but I'm not sure how to get an equation to maximize.

    I have to find it using different ways:
    a) using area as a function of the height,
    b) using area as a function of half the angle
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  2. #2
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    I don't need to do anything with the radius of 4?
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  3. #3
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    What I had done originally, trying to include the radius is:
    A = 1/2 (4+h)(4+h),
    A'= 1/2 (8 + 2h), but that gave me a negative critical number.
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  4. #4
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    I don't understand how to solve it the 2 different ways.
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  5. #5
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    Quote Originally Posted by janvdl View Post
    No we do need it, it sets the boundary for the triangle's size. I'm just not sure how to bring it into this situation.
    So I've gotta find help elsewhere?
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by unluckykc View Post
    So I've gotta find help elsewhere?
    Of course not. Someone will be along presently to help you with this. You'll have an answer soon.
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  7. #7
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    ok, thanks any way though
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  8. #8
    Bar0n janvdl's Avatar
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    A = \frac{1}{2} (x)(y)

    = \frac{1}{2} (x) ( \sqrt{ 4 - x^2 })

    A^2 = \frac{1}{4} x^2 (4 - x^2)

    = x^2 - \frac{1}{4} x^4

    A = (x) \sqrt{ (1 - \frac{1}{2} x) (1 + \frac{1}{2} x) }

    EDIT: This seems to be university work, you'll have to differentiate this part...
    Last edited by janvdl; November 14th 2007 at 01:05 AM.
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  9. #9
    Super Member angel.white's Avatar
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    SETUP
    Right, first set up your triangle like the one in the picture. Now lets set our domain for \theta. It will be easiest if we just take it through the first 90º, because beyond that things will be slightly confusing. We can see this doesn't create a problem, because it is blatantly obvious that no triangle who's points touch the circle above the center will be the biggest.

    So domain: 0\leq\theta\leq\frac{\pi}{2}

    Now we are looking for area, so lets write an equation to find our area.
    We see that we can split this triangle into two 90º triangles, solve for one, and then double that area. So first find the area for one of our 09º triangles:
    A=\frac{bh}{2} (note, we are talking about total height here, not the height I marked in my image)

    Now we want this for both triangles, so multiply it by 2
    A = bh
    So this is the equation for the area of our isosceles triangle.

    Edit: I should have used better terms, this is a bit confusing, this h is the total height, and this b is 1/2 the base. so this formua we are using is really 2 times the formula for the area of half of the triangle (twice one of the ninety degree triangles) That is why it may seem to not be the correct formula. Sorry, I'll keep that in mind next time.

    --------------

    WRITE AN EQUATION
    Now, we need to get those variables into something we can work with, we need a way to relate either base to height, or height to base, I chose to relate them to something they have in common. In this case, we'll relate them to \theta, which you can see is marked in the attached image.

    So now we need to solve the base from theta. Well we know the hypotenuse is equal to 4, and the base is the opposite side, so let's use sine.
    sin(\theta)=\frac{b}{4}

    Now solve for base
    4sin(\theta)=b

    And using the same methods, we can find the height from the center of the circle to the bottom of the triangle through
    cos(\theta)=\frac{h}{4}

    4cos(\theta)=h

    Excellent. Only problem is that this is the height from the center of the circle, we want it in terms of the entire circle. Well, we know that the distance from the center of the circle to the top of the circle is 4, so we can just add 4 to this answer to get the total height of the circle

    so
    4cos(\theta)+4=h

    Now substitute our values into our formula for area
    A=bh

    A=4sin(\theta)(4cos(\theta)+4)

    And simplify
    A=16sin(\theta)cos(\theta)+16sin(\theta)

    Now this equation will tell us the area of our whole isosceles triangle, based on the angle \theta

    ----------

    FIND THE MAXIMUM AREA
    So let's find the greatest area based on that angle. We know that if we graph this equation, it's highest point will be our highest area, and before and after that point, it will have a lower area. So it's slope is increasing up to that point, and decreasing after that point, and is equal to zero right at that point. So let's find where the slope is equal to zero.

    Well, we know that the derivative of this equation will give us the equation of the slope, then we can simply set the equation of the slope equal to zero, and we will know what that point is. So lets differentiate
    \frac{d}{dx}A=16(cos^{2}(\theta)-sin^{2}(\theta))+16cos(\theta)

    Now we set it equal to zero
    0=16(cos^{2}(\theta)-sin^{2}(\theta))+16cos(\theta)

    And solve:
    0=16(cos^{2}(\theta)-sin^{2}(\theta))+16cos(\theta)

    0=16cos^{2}(\theta)-16sin^{2}(\theta)+16cos(\theta)

    16sin^{2}(\theta)=16cos^{2}(\theta)+16cos(\theta)

    sin^{2}(\theta)=cos^{2}(\theta)+cos(\theta)

    1-cos^{2}(\theta)=cos^{2}(\theta)+cos(\theta)

    (1-cos(\theta))(1+cos(\theta))=cos^{2}(\theta)+cos(\t  heta)

    (1-cos(\theta))(1+cos(\theta))=cos(\theta)(cos(\theta  )+1)

    1-cos(\theta)=cos(\theta)

    1=2cos(\theta)

    \frac{1}{2}=cos(\theta)

    cos^{-1}(\frac{1}{2})=\theta

    Now, there are a number of values which solve this, but the only one within the domain that we chose is \frac{\pi}{3} So we know that when \theta=\frac{\pi}{3} that our tangent is equal to zero. Now, we need to check that this is in fact a maximum. Well, just looking at it we can see that it will be, but I chose to verify it by plugging in \frac{\pi}{3}+.01 and \frac{\pi}{3}-.01 and checking my area, they were each slightly smaller, so it is a max. You could also check that this is a maximum through your first derivative, or the best way to do it is through your second derivative, but I didn't wan to differentiate again, so I just did it the ghetto way.

    ---------

    CALCULATE THE AREA
    Now, we just need to plug in values to find our area.

    Our equation:
    A=16sin(\theta)cos(\theta)+16sin(\theta)

    Substitute \theta=\frac{\pi}{3}
    A=16sin(\frac{\pi}{3})cos(\frac{\pi}{3})+16sin(\fr  ac{\pi}{3})

    Use actual values:
    A=16\frac{\sqrt{3}}{2}\cdot\frac{1}{2}+16\frac{\sq  rt{3}}{2}

    Simplify
    A=4\sqrt{3}+8\sqrt{3}
    A=12\sqrt{3}

    So our maximum area equals 12\sqrt{3} square units, or about 20.7846 square units



    ...unless I messed up
    Attached Thumbnails Attached Thumbnails Maximum Area-circletriangle.jpg  
    Last edited by angel.white; November 14th 2007 at 01:39 AM.
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