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Math Help - practice worksheet on 2nd partials test

  1. #1
    Junior Member samsum's Avatar
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    practice worksheet on 2nd partials test

    find all the relative maxima, minima and saddle points of each of the following

    a) f(x,y)=3x^2+6xy+7y^2-2x+4y
    b) f(x,y)=x^3 - y^3 - 2xy+6
    c) f(x,y)=4xy-x^4-y^4

    problem 2) determine the dimensions of rectangular box, open at the top, having a volume of 32 cubic ft. Requiring the least amount of material for its construction.


    thanks for your help....these work out problem to practice for the exams. need your help...thanks
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  2. #2
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    Hello, samsum!

    I'll do part (c) . . .


    Find all the relative maxima, minima and saddle points.
    . . (c)\;f(x,y)\:=\:4xy-x^4-y^4

    Set the partial derivatives equal to zero and solve.
    . . \begin{array}{cc}f_x \:=\:4y - 4x^3\:=\:0 \quad\Rightarrow\quad y = x^3 \\ f_y \:=\:4x - 4y^3 \:=\:0\quad\Rightarrow\quad x = y^3\end{array} \begin{array}{cc}(1)\\(2)\end{array}

    Substitute (1) into (2): . x \:=\:\left(x^3\right)^3\quad\Rightarrow\quad x^9 - x \:=\:0\quad\Rightarrow\quad x(x^8-1) \:=\:0

    . . which has roots: . \begin{array}{ccc}x = 0,\;y = 0 \\x = 1,\;y = 1 \\ x = \text{-}1,\;y = \text{-}1\end{array}


    Second Derivative Test: . \begin{array}{ccc}f_{xx} = -12x^2\\ f_{yy} = -12y^2 \\ f_{xy} = 4\end{array}

    Then: . D \:=\:(f_{xx})(f_{yy}) - (f_{xy})^2 \:=\:(-12x^2)(-12y^2) - 4^2\:=\:16(9x^2y^2 - 1)


    At (0,0):\;D \:=\:16(0 - 1) \:=\:\text{-}16 . . . saddle point at (0,0,0)

    At (1,1):\;D \:=\:16(9-1) \:=\:+128 . . . minimum at (1,1,2)

    At (\text{-}1,\text{-}1):\;D \:=\:16(9-1) \:=\:+128 . . . minimum at (\text{-}1,\text{-}1,2)



    2) Determine the dimensions of an open-top rectangular box having a volume of 32 ft³
    requiring the least amount of material for its construction.

    Let x = length, y = width, z = height.

    The volume is: . xyz\,=\,32\quad\Rightarrow\quad z \,=\,\frac{32}{xy} [1]

    The surface area is: . A \;= \;xy + 2yz + 2xz [2]

    Substitute [1] into [2]: . A\;=\;xy + 2y\left(\frac{32}{xy}\right) + 2z\left(\frac{32}{xy}\right)

    . . and we have: . A \;= \;xy + 64y^{-1} + 64x^{-1} . . . which we will minimize.


    Set the partial derivatives equal to zero and solve.

    . . \begin{array}{cc}A_x  \:= \:y - 64x^{-2} \:=\:0 \\ A_y\:=\:x-64y^{-2}\:=\:0\end{array}\quad \begin{array}{cc}\Rightarrow \\ \Rightarrow\end{array}\quad \begin{array}{cc}y \:=\:\frac{64}{x^2}\\ xy^2\:=\:64\end{array} \begin{array}{cc}(3)\\(4)\end{array}<br />

    Substitute (3) into (4): . x\left(\frac{64}{x^2}\right)^2\:=\:64\quad\Rightar  row\quad x^3 = 64\quad\Rightarrow\quad\boxed{ x = 4}

    Substitute into (3): . y \:=\:\frac{64}{4^2}\quad\Rightarrow\quad\boxed{y = 4}

    Substitute into (1): . z \:=\:\frac{32}{4\cdot4}\quad\Rightarrow\quad\boxed  {z = 2}


    Therefore, the dimensions of the box are: .  4 \times 4 \times 2 feet.

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