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Math Help - min/max for partial differential

  1. #1
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    min/max for partial differential

    U = x  + 2xy + y

    What's the minimum or maximum of U?

    Differentiating I get:

    dU = (1 + 2y)dx + (2x + 1)dy

    So, is it correct to say that both the partial differentials are constant, therefore there is no minimum or maximum?

    -Scott
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  2. #2
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    Hello, Scott!

    U(x,y) \:= \:x  + 2xy + y

    What's the minimum or maximum of U ?
    We're expected to equate the partial derivaitves to zero, and solve.

    . . U_x \:=\:1 + 2y \:=\:0\quad\Rightarrow\quad y \:=\:-\frac{1}{2}

    . . U_y \:=\:2x+1\:=\:0\quad\Rightarrow\quad x \:=\:-\frac{1}{2}

    U\left(\text{-}\frac{1}{2},\,\text{-}\frac{1}{2}\right) \:=\:-\frac{1}{2} + 2\left(\text{-}\frac{1}{2}\right)\left(\text{-}\frac{1}{2}\right) - \frac{1}{2} \:=\:-\frac{1}{2}


    Since I could see that U(x,y) can increase without bound,
    . . I suspected that we had found a minimum value.

    Then I tried the Second Partials Test . . .

    . . . . U_{xx} \:=\:0\qquad U_{yy} \:=\:0 \qquad U_{xy} \:=\:U_{yx} \:=\:2

    D \;=\;\left(U_{xx}\right)\left(U_{yy}\right) - \left(U_{xy}\right)^2 \;=\;0\cdot0 - 2^2\;=\;-4 . . . . negative


    Therefore: . \left(-\frac{1}{2},\:-\frac{1}{2},\:-\frac{1}{2}\right) .is a saddle point.

    . . The function has neither a maximum nor a minimum.

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