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Math Help - help with two problems

  1. #1
    Newbie ayato's Avatar
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    help with two problems

    Hi, i have two problems that i really need help with. The problems are due tomorrow at 2:30. can someone please help me out, i'll really appreciate it.


    implicit differentiation problem
    1) (x^2+y^2)/(x+y)=xy-2
    find derivitive (dy/dx) at point (-1, -1)

    maxima/minima problem
    2)a cylinder, including top and bottom, is madde from material which costs "x" dollars per square inch. Suppose their is an additional cost of fabrication given by "n" dollars per inch of the circumfrence of the top and bottom of the can. find an algerbraic equation whose solution would be he radius , r, of the can of given volume, V, whose cost is a minimum.
    * Do not have to solve equation, just find a algerbraic solution

    Thank you in advance.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ayato View Post
    implicit differentiation problem
    1) (x^2+y^2)/(x+y)=xy-2
    find derivitive (dy/dx) at point (-1, -1)
    x^2 + y^2 = (xy - 2)(x + y)

    x^2 + y^2 = x^2 y + xy^2 - 2x - 2y

    Now take the derivative:
    2x + 2yy' = 2xy + x^2 y' + y^2 + 2xyy' - 2 - 2y'

    2yy' - x^2 y' - 2xyy' + 2y' = -2x + 2xy + y^2 - 2

    y'(2y - x^2 - 2xy + 2) = y^2 + 2xy - 2x - 2

    y' = (y^2 + 2xy - 2x - 2)/(2y - x^2 - 2xy + 2)

    Now, we are at the point (-1, -1):
    y' = ((-1)^2 + 2(-1)(-1) - 2(-1) - 2)/(2(-1) - (-1)^2 - 2(-1)(-1) + 2)

    y' = (1 + 2 + 2 - 2)/(-2 - 1 - 2 + 2)

    y' = 3/-3 = -1

    -Dan
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  3. #3
    Newbie ayato's Avatar
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    Thanks a ton man. I never thought about getting rid of the fraction from the beginning, only trying to get rid of it after I differentiated. This, as you can prolly tell, made the problem more complex than it should have been.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ayato View Post
    maxima/minima problem
    2)a cylinder, including top and bottom, is madde from material which costs "x" dollars per square inch. Suppose their is an additional cost of fabrication given by "n" dollars per inch of the circumfrence of the top and bottom of the can. find an algerbraic equation whose solution would be he radius , r, of the can of given volume, V, whose cost is a minimum.
    * Do not have to solve equation, just find a algerbraic solution
    The surface area of a cylinder is:
    S = 2(pi)r^2 + 2(pi)rL, where r is the radius of the end cap and L is the length of the cylinder.

    The volume of a cylinder is
    V = (pi)r^2 L is considered to be a constant.

    So the cost function will be:
    C = x[2(pi)r^2 + 2(pi)rL] + n(2)[2(pi)r]

    C = 2(pi)rx[r + L] + 4n(pi)r

    Now, let's find dC:

    dC = (DC/Dr)dr + (DC/DL)dL (D/Dr is the first partial derivative with respect to r.)

    dC = {2(pi)x[r + L] + 2(pi)rx + 4n(pi)}dr + {2(pi)rx}dL

    Now, we may find a relationship between dr and dL by using the differential of the volume equation (recall that V is constant!):
    0 = {2(pi)rL}dr + {2(pi)r^2}dL

    dL = -{[2(pi)rL]/[2(pi)r^2]}dr = -{L/r}dr

    So
    dC = {2(pi)x[r + L] + 2(pi)rx + 4n(pi)}dr + {2(pi)rx}*-{L/r}dr

    dC = {2(pi)x[r + L] + 2(pi)rx + 4n(pi) - 2(pi)Lx}dr

    Thus
    dC/dr = 2(pi)x[r + L] + 2(pi)rx + 4n(pi) - 2(pi)Lx

    dC/dr = 2(pi)xr + 2(pi)xL + 2(pi)rx + 4n(pi) - 2(pi)Lx

    dC/dr = 4(pi)rx + 4n(pi)

    (Notice that this no longer depends on L. If it did we would have to use the volume equation to remove it.)

    Now set this equal to 0 and find r. (You should then verify that this r value does indeed give you a minimum cost instead of a maximum.)

    -Dan
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  5. #5
    Newbie ayato's Avatar
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    Thank you very much for the help with the cylinder question. I had the formula correct, but was unsure of how to get it into the form of just r and x and n.

    Now I realize you have to prove this is a minimum value, but Im not sure how to do this. I know the graph of this will a parabola which is concave up meaning there is a minimum value, but how do you prove this equation gives you a min. value?

    thanks again and sorry for all the questions!
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ayato View Post
    Thank you very much for the help with the cylinder question. I had the formula correct, but was unsure of how to get it into the form of just r and x and n.

    Now I realize you have to prove this is a minimum value, but Im not sure how to do this. I know the graph of this will a parabola which is concave up meaning there is a minimum value, but how do you prove this equation gives you a min. value?

    thanks again and sorry for all the questions!
    I live to serve.

    Use the second derivative test:

    dC/dr = 4(pi)rx + 4n(pi)

    d^2C/dr^2 = 4(pi)x

    Since the second derivative is positive at your r value (actually for ANY r value) the r that you found will be a minimum value for the function.

    -Dan
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  7. #7
    Newbie ayato's Avatar
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    Thanks a ton, you saved me!! I forgot all about the second deriv. test (amazing how we learn it in class and never think about it) That is an excellent method, thank you very much
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ayato View Post
    Thanks a ton, you saved me!! I forgot all about the second deriv. test (amazing how we learn it in class and never think about it) That is an excellent method, thank you very much
    You are welcome!

    -Dan
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