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Math Help - partial differentiation

  1. #1
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    partial differentiation

    Hello

    I have two functions:

    V(x,h)=x^2 \cdot h=4

    and

    A(x,h)=2x^2+4xh

    As someone might guess, this is a optimization problem of sorts (minimization of A(x,h) with V(x,h)=4 as constraint). My problem arises when I attempt doing partial differentiation, and the root of the problem is perhaps that I have forgotten something.

    When, for instance, I take the partial derivatives of A(x,h), I get this:

    A_x=4x+4h \\  A_h=4x

    It doesn't seem right, and I think that the problem is the thing with variables multiplied by eachother. Can someone please point me in the right direction with this?

    Last edited by JacobOCD; January 17th 2013 at 08:06 AM.
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  2. #2
    Junior Member Barioth's Avatar
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    Re: partial differentiation

    Hi you derivate are good. Maybe if you explain the question you trying to solve I can help more.

    Try the Lagrange multiplier?

    Note if you want to remove the <br/>, dont hit enter in you latex code. if you want your code to be swapping line write \\
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    Re: partial differentiation

    Well, it might seem silly asking for a solution to my problem, but that is what actually would help, so I can compare it with the solutions for other problems and draw some conclusions as to method from there.

    As I wrote earlier, my problem is about minimizing

    A(x,h)=2x^2+4xh

    as long as the solution fulfills

    V(x,h)=x^2 \cdot h=4

    It must be mentioned that I haven't had any relevant calculus courses to actually solve the problem.
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  4. #4
    Junior Member Barioth's Avatar
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    Re: partial differentiation

    I'll guess you don't know about lagrange multiplier and work without them (anyway this is probably faster without it!)
    aight in V(x,h) = h = \frac{4}{(x^2)}

    plug in the A(x,h) = 2x^2+\frac{4x*4}{x^2}=2x^2+\frac{16}{x^2}=f(x)

    Now we need to find the min of f(x). This you should be able to!

    \frac{d(f(x))}{dx}= 4x +\frac{-32}{x^3}

    if f'(x) = 0

    then x= +- 2^{\frac{3}{4}

    these iare the only point that might be a min or a max. (I'll leave to you to check if it's a min or a max. and its value.)

    edit: Typo mistake, I need to slow down when I type stuff..

    hope that helped
    Last edited by Barioth; January 17th 2013 at 08:22 AM.
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    Re: partial differentiation

    Well, that was actually my first approach. What happened was that I got some quite weird results, but since you again propose the same method (which would imply that it might actually work), I will try redoing it, and I will see what I get. I might have made some algebraic screw-up the first time i did it. I'll post, once I see if it works, but I do recognise 2^\frac 2 3 from my own solution.
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  6. #6
    Junior Member Barioth's Avatar
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    Re: partial differentiation

    its 2^(3/4) not 2^(2/3) watch out!
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    Re: partial differentiation

    BTW, err... From my solution:

    A(x,h) with h defined and substituted:

    A(x)=2x^2+4x \cdot \frac {4}{x^2} = 2x^2 + \frac {16x} {x^2} = 2x^2 + \frac {16}{x}

    A'(x)=4x - \frac{16}{x^2}

    Solve the critical point equation:

    A'(x)=4x - \frac{16}{x^2}=0
    x=2^\frac{2}{3}

    Same result as earlier :/

    BTW:

    A(2^\frac{2}{3}, h) \neq A(2^\frac{2}{3}) ...

    My reason for suspicion. The h has obviously been found by solving the constraint equation for h with 2^\frac{2}{3} substituted for x.
    Last edited by JacobOCD; January 17th 2013 at 08:39 AM.
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  8. #8
    Junior Member Barioth's Avatar
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    Re: partial differentiation

    Watch out I made a mistake when I sustitued the h. but the answer you have there seem good. I'm in my class right now. I'll check this out during the break
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  9. #9
    Junior Member Barioth's Avatar
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    Re: partial differentiation

    To find H put that value on X in V(x,h).

    your answer seem very good to me!

    still not sure? Let me know I'll try to be more precise.
    Last edited by Barioth; January 17th 2013 at 11:06 AM.
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    Re: partial differentiation

    I spent some time researching Lagrange multipliers, and I somewhat got the idea of it, but I have been making some sort of mistake when attempting to apply the method, in the sense that I could not solve the three equations, it yielded ... which might imply errors in terms of both algebra and calculus. There is quite a few places, I might have slipped up , so if it's not too time consuming (or actually, it is quite time consuming using Lagrange multipliers, it seems), I would appreciate an example of a solution by that method for reasons of comparison.
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  11. #11
    Junior Member Barioth's Avatar
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    Re: partial differentiation

    We have  \\A(x,h)=2x^2+4xh\\ V(x,h)=x^2h=4

    we start by evaluating the derivate.

     \\V_x =2xh\\V_h=x^2\\A_x=4x+3h\\A_h=4x\\
    Now that we know the derivate we use the langrage multiplier we end up with these equation
    \\A_x=\lambda V_x\Rightarrow 4x+3h = \lambda (2xh)\\A_h=\lambda V_h\Rightarrow 4x = \lambda (x^2)\\V(x,h)=x^2h=4

    You then have to find the value of x,h and \lambda. you should end with x = 2^{\frac{2}{3}}

    thats pretty much the idea of it. Both method work, its up to you to find the "Best" one for each problem. but you should not have different answer.
    Last edited by Barioth; January 17th 2013 at 11:39 AM.
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    Re: partial differentiation

    That is actually very similar to what I did. I will go to bed soon, but I'll try redoing it by method of Lagrange tomorrow, and we'll see what happens
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  13. #13
    Junior Member Barioth's Avatar
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    Re: partial differentiation

    Aight let me know how it ended!
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