What Is The Geometric Meaning Of Third Derivative Of A Function At A Point?

Help finding critical point of a function of two variables?

• I need to find the critical points of this function: f(x,y)=(12x-x^2)(10y-y^2). There are five points and I've found four but I'm not sure how to find the last one. The ones I have are (0,0), (0,10), (6,5), and (12,0). I also need help figuring out if the third point is a minimum, maximum, saddle point or neither. I've done the second derivative test a couple times and keep getting saddle point, which is wrong. Any help would be greatly appreciated. Thanks

• Answer:

  f(x,y)=(12x-x^2)(10y-y^2) The critical points of a continuous function are the points where all partial derivatives are equal to zero. In this case, we find the following partial derivatives: ∂f/∂x = (12 - 2x)(10y-y^2) ∂f/∂y = (12x-x^2)(10 - 2y) The first is zero when: x = 6 OR y = 0 OR y = 10 The second is zero when: x = 0 OR x = 12 OR y = 5 So, in order to find all critical points, you need to pair up each compatible set of conditions. The critical points are then: (6,5), (0,0), (12,0), (0,10), (12,10) (the last one was the one you were missing) In order to find the critical points, you need to work with the second partial-derivative matrix, aka its Hessian. More on that when I can. We now find the successive partial derivatives: ∂^2f/∂x^2 = -2*(10y-y^2) ∂^2f/∂x∂y = ∂^2f/∂y∂x = (12 - 2x)(10 - 2y) ∂^2f/∂y^2 = -2(12x - x^2) At each critical point, we consider the matrix H = {{∂^2f/∂x^2, ∂^2f/∂x∂y}, {∂^2f/∂y∂x, ∂^2f/∂y^2}} If the determinant of the matrix is positive and ∂^2f/∂x^2 (the upper-left entry) is positive, then the critical point is a minimum If the determinant of the matrix is positive and ∂^2f/∂x^2 (the upper-left entry) is negative, then the critical point is a maximum If the determinant of the matrix is negative, then the critical point is a saddle point If the determinant of the matrix is zero, then the test is inconclusive. Now to use this test at each point: (0,0) yields {{0, 120},{120, 0}} The determinant of this matrix is 0*0 - 120*120 = -14400, which is negative. So, (0,0) is a saddle point (0,10): {{0,-120},{-120,0}} The determinant is -14400, so (0,10) is a saddle point (6,5): {{-50, 0},{0, -72}} The determinant is 3600, which is positive, and the upper left entry is -50, which is negative, so the critical point is a maximum (12,0): {{0, -120},{-120, 0} Saddle Point (12,10): {{0, 120},{120, 0}} Saddle Point\