Find and classify all local min and maxima and saddle points of the function f(x,y)=/3yx^2-3xy^2+36xy
I know there are 3 saddle points and one maxima.
This is what I got:
D=FxxFyy-(Fxy)^2
= 36xy-36(x+y-6)^2
But how do i solve for zero?? Im lost on this one. Please help me figure out these points??
First you'd have to find the local maxima/minima. If you have obtained the (four) points, then you can use the second derivative test to determine if each one is a maximum/minimum or saddle point.
Have you found the critical points?
(Unfortunately the definition of f(x,y) above does not seem to be complete.)
Note:singular: maximum, plural: maxima.
Lucy
I think you are wrong,what I got is much that I can't type ,mayb some social network or video call .
To find the critical points of a function, you need to determine where the gradient is zero or undefined. In this case, we have a function of two variables f(x, y) = 3yx^2 - 3xy^2 + 36xy.
To find the critical points, we need to find where the gradient is zero. The gradient of a function is given by the partial derivatives with respect to each variable. So, let's find the partial derivatives of f(x, y) with respect to x and y:
∂f/∂x = 6yx - 3y^2 + 36y
∂f/∂y = 3x^2 - 6xy + 36x
To find the critical points, set both partial derivatives equal to zero and solve the resulting system of equations:
6yx - 3y^2 + 36y = 0 (equation 1)
3x^2 - 6xy + 36x = 0 (equation 2)
Now, you can try to solve this system of equations by substitution or elimination. However, it seems you have already calculated the determinant (D) of the Hessian matrix, which will help us classify the critical points.
The Hessian matrix is a square matrix of second-order partial derivatives. For our function f(x, y), the Hessian matrix is:
H = |∂^2f/∂x^2 ∂^2f/∂x∂y|
|∂^2f/∂y∂x ∂^2f/∂y^2|
The determinant of the Hessian matrix, denoted as D, can be used to classify the critical points:
D = (∂^2f/∂x^2) * (∂^2f/∂y^2) - (∂^2f/∂x∂y)^2
In our case, you have correctly found D to be 36xy - 36(x + y - 6)^2.
To determine the nature of the critical points, we need to evaluate D at each critical point. If D > 0 and (∂^2f/∂x^2) > 0, then it is a local minimum. If D > 0 and (∂^2f/∂x^2) < 0, then it is a local maximum. If D < 0, then it is a saddle point.
To determine where D = 0, we evaluate the expression 36xy - 36(x + y - 6)^2 and set it equal to zero:
36xy - 36(x + y - 6)^2 = 0
Solving this equation will give you additional critical points to consider.