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Maxima, Minima, Saddle Point

  1. Apr 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the local max, min, and saddle point for the function:
    f(x,y) = 2x^2+3xy+4y^2-5x+2y


    2. Relevant equations


    3. The attempt at a solution
    I've taken the two partial derivatives

    Fx = 4x + 3y - 5
    Fy = 3x + 8y + 2

    I know that the critical points will sit where both of theses partial derivatives = 0
    i.e.

    Fx = 4x + 3y - 5 = 0
    Fy = 3x + 8y + 2 = 0

    The problem I have here though is that I don't know how to solve the system of equations.

    I know once I've solved the system of equations I can use the determinant of the jacobian matrix to see whether they are local max, min, or saddle points...

    Any help with solving the system of equations would be much appreciated. I've had a bit of trouble solving systems of equations in the past.
     
  2. jcsd
  3. Apr 2, 2016 #2

    SteamKing

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    Really? You never solved a set of simultaneous linear equations in your algebra courses?

    You can use Cramer's Rule or elimination to solve the system above.

    http://www.coolmath.com/algebra/14-determinants-cramers-rule/01-determinants-cramers-rule-2x2-01

    http://www.purplemath.com/modules/systlin6.htm

     
  4. Apr 3, 2016 #3

    HallsofIvy

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    Generally speaking, it is not a good idea to try to learn Calculus until after you have a firm grasp of algebra. You have the equations
    Fx = 4x + 3y - 5 = 0 and Fy = 3x + 8y + 2 = 0. If you multiply the first equation by 3 you have 12x+ 9y- 15= 0. If you multiply the second equation by 4 you have 12x+ 32y+ 8= 0. Now the x term in each equation has the same coefficient so subtracting one equation from the other eliminates x and you have a single equation to solve for y.
     
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