Find the critical points and determine whether each is a Relative Max, Relative Min, or Saddle Point.

Question

the function f(x,y) = 2x^4 + y^2 - 12xy has derivatives, fx = 8x^3-12y, fy = 2y-12x

 Find the critical points and determine whether each is a Relative Max, Relative Min, or Saddle Point.

 

Just wanted to know the right answer. i got it wrong on my quiz

Answer 1

   0 = fx = 8x^3 - 12y

 

  0 = fy = 2y - 12x -->   12x = 2y  -->    y = 6x

  plugging this into the first equation

 

  0 = 8x^3 - 12(6x) = 8x^3 - 72x

           0 = 8x( x^2 - 9)

          0 = 8x(x+3)(x-3)

      x=0  ,  x = 3,    x=-3

 

    (0,0) (3,18)  (-3,-18)  <-- critical points

 

 

Now we need the second partial derivatives:

  fxx = 24x^2 

 fyy = 2

 fxy = -12

 

 

D (a,b) = fxx(a,b)*fyy(a,b) - fxy(a,b)^2

              

1st critical point (0,0):       fxx(0,0) *fyy(0,0) - fxy(0,0)^2 = 0*0 - (-12)^2 = 0 - 144 < 0.

                                           So (0,0) is a saddle point

 

2nd critical point (3,18):   fxx(3,18)*fyy(3,18) - fxy(3,18)^2 = 24*3^2 * 2 - (-12)^2 =

                                                                                                 24*9*2 - 144 =

 

                                                                                                  432-144 = 288

                                           So (3,18) is a relative minimum because D>0 and fxx>0

 

3rd critical point (-3,-18): fxx(-3,-18)*fyy(-3,-18) - fxy(-3,-18)^2 = 24*(-3)^2 * 2 - (-12)^2 =

                                                                                                       24*9*2 - 144 =

                                                                                                         432 - 144 = 288

  So (-3,-18) is also a relative minimum because D>0 and fxx > 0

 

SOURCE:

Paul's online math notes

http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx