Unfortunately that is the entire problem..

Find and classify all local minima, local max, and saddle points of the function F(x,y)= -3x^2y-3xy^2+36xy

There should be 3 saddle points and a maxima for sure! Please help me get the points. I can't do these maxima minima problems. they are confusing as heck! Im about to have a test over this, freaking out!!

F(x,y)=-3x^2y-3xy^2+36xy

For maximum/minimum,Fx=0 AND Fy=0
Fx=36y-6xy-3y^2=3y(12-2x-y)=0 ...(1)
Fy=36x-6xy-3x^2=3x(12-2y-x)=0 ...(2)

Clearly (0,0) is a critical point.
If y=0, solve for x in (2) to get x=12, so
(12,0) is another critical point.
Similarly (0,12) is another critical point.
Finally, solve for x,y in
12-2x-y=0 and
12-x-2y=0 to get
x=y=4, or (4,4) is a critical point.

So the four critical points are
(0,0)
(0,12)
(12,0)
(4,4)

Second derivative test
Calculate
Fxx=-6y
Fyy=-6x
Fxy=Fyx=6(6-x-y) [Clairot's theorme]
Now calculate
D(x,y)=Fxx.Fyy-Fxy²
=36xy-6(36-12(x+y)+(x+y)^2)
D(0,0)=-216<0 => saddle point
D(0,12)=-216<0 => saddle point
D(12,0)=-216<0 => saddle point
D(4,4)=552 >0 => max or min
Fxx=-6(4)=-24 <0 => maximum

I understand that finding local minima, local maxima, and saddle points can be challenging. Don't worry, I'll explain the steps to help you solve this problem.

To find the local minima, local maxima, and saddle points of the function F(x, y) = -3x^2y - 3xy^2 + 36xy, we need to follow these steps:

Step 1: Calculate the first-order partial derivatives with respect to x and y.
Step 2: Set both partial derivatives to zero and solve the resulting system of equations to find the critical points.
Step 3: Calculate the second-order partial derivatives with respect to x and y.
Step 4: Use the second-order partial derivatives to determine the nature of each critical point.

Now, let's go step by step:

Step 1: Calculate the first-order partial derivatives with respect to x and y.
To find the partial derivative with respect to x, we differentiate F(x, y) with respect to x while treating y as a constant:
Fₓ = -6xy - 3y² + 36y.

To find the partial derivative with respect to y, we differentiate F(x, y) with respect to y while treating x as a constant:
Fᵧ = -3x² - 6xy + 36x.

Step 2: Set both partial derivatives to zero and solve the resulting system of equations to find the critical points.
To find the critical points, we'll set both Fₓ and Fᵧ equal to zero and solve the resulting equations simultaneously:

-6xy - 3y² + 36y = 0 ...(Equation 1)
-3x² - 6xy + 36x = 0 ...(Equation 2)

Step 3: Calculate the second-order partial derivatives with respect to x and y.
To find the second-order partial derivatives, we differentiate each of the first-order partial derivatives found in Step 1.

To find the second-order partial derivative with respect to x, we differentiate Fₓ with respect to x:
Fₓₓ = -6y.

To find the second-order partial derivative with respect to y, we differentiate Fᵧ with respect to y:
Fᵧᵧ = -6x.

Step 4: Use the second-order partial derivatives to determine the nature of each critical point.
To determine the nature of each critical point (whether it's a local minimum, local maximum, or saddle point), we'll use the second-order partial derivatives and the discriminant (D = Fₓₓ * Fᵧᵧ - (Fₓᵧ)²) at each critical point.

For each critical point (x₀, y₀), calculate:
D = Fₓₓ(x₀, y₀) * Fᵧᵧ(x₀, y₀) - (Fₓᵧ(x₀, y₀))².

If D > 0 and Fₓₓ(x₀, y₀) > 0, then it is a local minimum.
If D > 0 and Fₓₓ(x₀, y₀) < 0, then it is a local maximum.
If D < 0, then it is a saddle point.
If D = 0, the test is inconclusive.

By following these steps, you can determine the nature of the critical points and classify them as local minima, local maxima, or saddle points.