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Math Help - Need help with Maximum/Minimum Problems...

  1. #1
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    Unhappy Need help with Maximum/Minimum Problems...

    I'm having some issues with solving these 2 problems.
    The first one is as follows:

    Find the value of a for which the area of the triangle formed by the tangent (y=a^2 - 2ax + 4) and the co-ordinate axes will be minimum.

    I calculated the distances of the side and base of the triangle by finding the intercepts of the tangent.
    Heres what I got: side = (a^2 + 4)/2 and base = a^2 + 4

    Therefore, my main equation for the area was 1/2*side*base

    I simplified the area equation, differentiated it (using quotient rule) and got a= +2 or -2

    I tested both with the first derivative, and found a=2 produced the local max.
    Therefore, the area would be 8 units squared

    However the answers say the area is 2/3*\sqrt{3}

    The second question is similar:

    Find the maximum area of a right triangle with a hypotenuse of 16cm

    my starting equation was y^2 = \sqrt{256-x^2}

    subbed into the area equation and differentiated, i reached a dead end...i think i ended up with the first derivative test screwing up...

    the answer is apparently 64cm^2

    Any help on these 2 questions? Ive been working on them for ages and keep repeating the same steps to no avail...

    Thanks!
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  2. #2
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    I calculated the distances of the side and base of the triangle by finding the intercepts of the tangent.
    Heres what I got: side = (a^2 + 4)/2 and base = a^2 + 4
    Look at the equation you have for the side again. Setting x=0 gives you  y=a^2+4 but setting y=0 does gives x=\frac{a^2+4}{2a}

    For the second question, you have x^2+y^2=16^2 and you want \frac{1}{2}xy to be maximum. So sub in y=\sqrt{256-x^2}, diffrentiate to get what x you have, and work back to get the area( hint, the x and the y are going to be the same)
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  3. #3
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    Oops, i accidentally switched the intercept values when i wrote out the question. they were right when i tried solving the problem on paper...so im still stuck >_<

    As for the second one, i tried differentiating this time with the product rule.

    heres what i got

    A' = (x/2).(1/4\sqrt{256-x^2}) + (1/2)(\sqrt{256-x^2}/2)
    That becomes:
    (x+512-2x^2)/8\sqrt{256-x^2}

    And again i run into the same problem

    For stat. points, x + 512 - 2x^2 = 0
    But when i factorise that using the quadratic formula for 0 i get a figure with a surd in it...
    Im pretty sure if i continue with it it wont produce the 64cm^2 that is the answer

    so what did i do wrong?
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  4. #4
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    For Problem A:

    My intercept value is not the same as yours for the side, you are missing an a. So your value for the Area is \frac{(a^2-4)^2}{2a}. Is this the value you had? The diffriation for this is quite complicated as well, if you want to see a sample diffrentiation try:http://www2.wolframalpha.com/Calcula...=3&w=461&h=874

    For problem B:

    The answer you are looking for will have a surd in it. Remember that the value the equation gives you is the value for x, not the value for the area. To get the cvalue for the area (th 64 cm^2) you have to work out your y (from Y^2+X^2=256 and then get your area from area =\frac{1}{2}xy

    With regards to your diffrentiation itself:
     A=\frac{x}{2}\sqrt{256-x^2}
    chain Rule: with u=x/2 and v=\sqrt{256-x^2}
    A'=u\frac{dv}{dx} +v\frac{du}{dx}
    To get the \frac{dv}{dx} you use the product rule.
    \frac{dv}{dx}=\frac{dv}{db}\frac{db}{dx} where b=256-x^2, and v=\sqrt{b} (Do you how this works?)

    which leaves A'=\frac{x}{2}\times\frac{-2x}{2\sqrt{256-x^2}}+\sqrt{256-x^2}\times\frac{1}{2}
    Simplifying gives
    A'=\frac{128-x^2}{2\sqrt{256-x^2}}

    You only care when A'=0, so this simplifies to 0=128-x^2.

    Finally, to get the area you plug your x and y obtained above into A=0.5xy. That help?
    Last edited by Ackbeet; May 21st 2011 at 03:31 AM. Reason: Fixed incorrect LaTeX code.
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  5. #5
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    Thanks! I get problem b now, but shouldnt the area equation for prob A = (a^2 + 4)^2/4a ? since you have to multiply by half?
    thanks again
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  6. #6
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    Yes, it should be, sorry.
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