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    1. Given f(x,y) = x^2-4xy+y^3+4y
    Find the critical points and then use the Saddle Point Derivative Test to determine if they are max, min, or saddle points.

    2. Given f(x,y)=4xy-x^4-y^4
    Find the critical points and then use the Saddle Point Derivative Test to determine if they are max, min or saddle points.

    3. Find the volume of the solid bounded above by z=x^2 and abounded below by the region enclosed by y=2-x^2 and y=x.

    4. Find the volume of the solid in the first octant bounded by the coordinate planes z=4-y^2 and the plane x=3.

    5. Find the volume of the wedge cut from the first octant by the cylinder
    z=12-3y^2 and the plane x+y=2.

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    https://brainmass.com/math/complex-analysis/calculus-critical-point-derivatives-63711

    Solution Preview

    Please see the attached file.

    1. Given f(x,y) = x^2-4xy+y^3+4y
    Find the critical points and then use the Saddle Point Derivative Test to determine if they are max, min, or saddle points.
    2. Given f(x,y)=4xy-x^4-y^4
    Find the critical points and then use the Saddle Point Derivative Test to determine if they are max, min or saddle points.
    3. Find the volume of the solid bounded above by z=x^2 and abounded below by the region enclosed by y=2-x^2 and y=x.
    4. Find the volume of the solid in the first octant bounded by the coordinate planes z=4-y^2 and the plane x=3.
    5. Find the volume of the wedge cut from the first octant by the cylinder
    z=12-3y^2 and the plane x+y=2.

    1. Given f(x,y) = x^2-4xy+y^3+4y
    Find the critical points and then use the Saddle Point Derivative Test to determine if they are max, min, or saddle points.

    Solution:

    Here we need to find the critical points and maximum and minimum of the function

    Now let us come to our problem.

    Given f(x,y) = x^2-4xy+y^3+4y

    fx= 2x- 4y

    and fy= -4x+3y^2+4

    To find the critical points we have to make fx=0 and fy=0

    2x-4y=0 and -4x+3y^2+4 =0

    x= 2y ------ 1
    and x = 3/4(y^2) +1 ---------2

    From 1 and 2 we get,

    2y = ¾ (y^2)+1

    2y = 0.75y^2+1

    0.75y^2 -2y +1 =0

    The quadratic equation
    a x2 + b x + c = 0
    can always be solved by substituting the coefficients a , b and c into the quadratic formula :

    ________________________________________ ________________________________________
    -b ± b2 - 4 a c
    x = ________________________________________ ________________________________________
    2 a
    To make the following working a bit clearer, we can rewrite the equation
    0.75 x2 - 2 x + 1 = 0
    as
    0.75 x2 + (-2) x + 1 = 0.
    Comparing ...

    Solution Summary

    This solution is comprised of a detailed explanation to find the critical points and then use the Saddle Point Derivative Test to determine if they are max, min, or saddle points.

    $2.49

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