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Math Help - maximum percentage error

  1. #1
    Member billym's Avatar
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    maximum percentage error

    I have this function again:

    <br />
f(x,y) = x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6<br />

    The value of f when x = -1 and y = 2 is 25.

    The maximum percentage error in the value of x is 2 % and the maximum percentage error in the value of y is 3%.

    I have to find the least and greatest possible values that f(-1,2) can take, and the maximum possible percentage error in taking the value f(-1,2) = 25.

    I plugged -1 and 2 into the partial derivatives:

    <br />
fx (-1,2) = 4(-1)[(-1)^2 + (2)^2 + 1] = -24<br />

    <br />
fy (-1,2) = 4(2)[(2)^2 + (-1)^2 - 1] = 32<br />

    And then found:

    <br />
\delta f = -24(0.02)(-1) + 32(0.03)(2) = 2.4<br />

    So the least and greatest possible values for f(-1,2) are 22.6 and 27.4, and the maximum possible percentage error is 9.6% ?

    Is there anything correct here or am I completely wrong?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by billym View Post
    I have this function again:

    <br />
f(x,y) = x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6<br />

    The value of f when x = -1 and y = 2 is 25.

    The maximum percentage error in the value of x is 2 % and the maximum percentage error in the value of y is 3%.

    I have to find the least and greatest possible values that f(-1,2) can take, and the maximum possible percentage error in taking the value f(-1,2) = 25.

    I plugged -1 and 2 into the partial derivatives:

    <br />
fx (-1,2) = 4(-1)[(-1)^2 + (2)^2 + 1] = -24<br />

    <br />
fy (-1,2) = 4(2)[(2)^2 + (-1)^2 - 1] = 32<br />

    And then found:

    <br />
\delta f = -24(0.02)(-1) + 32(0.03)(2) = 2.4<br />

    So the least and greatest possible values for f(-1,2) are 22.6 and 27.4, and the maximum possible percentage error is 9.6% ?

    Is there anything correct here or am I completely wrong?
    What is wrong with your method is that you are approximating the change in the function by using the derivative at (-1, 2), but the function values you want to calculate are not at (-1, 2).

    I would find out if the derivative is equal to zero on the area of interest and the simply use the Extreme Value Theorem with endpoints x = -1.03, x = -.97, y = 1.96, y = 2.04. The minimum and maximum values of the function on that area must either occur where the derivative is zero or on the border.
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  3. #3
    Member billym's Avatar
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    The question says I have to use first-order partial derivatives to determine the values, but I can't find a single example of this kind of question in my notes.

    I have no idea what I am doing.



    Do 24.52 and 25.48 make any more sense?
    Last edited by billym; May 5th 2008 at 01:28 PM.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by billym View Post
    The question says I have to use first-order partial derivatives to determine the values, but I can't find a single example of this kind of question in my notes.

    I have no idea what I am doing.



    Do 24.52 and 25.48 make any more sense?
    I can't think of what you would be using the partial derivatives for except checking to see if \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0 at some point in the area of interest.
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