Find and classify all local minima, local maxima, and saddle points of the function f(x,y)= -3yx^2-3xy^2+36xy
I really need help answering this one!! Please answer in detail! Thank you
This is what I have so far...
D=FxxFyy-(Fxy)^2=(-6y)(-6x)-(-6x-6y+36)^2
= 36xy-36(x+y-6)^2
Now set D=0 but I don't know how to solve it?? 36xy-36(x+y-6^2=0
Set Fx=0=-6x-6y+36 and set Fy=0=-6xy-3y^2+36y
I need help with the algebra, its been so long
To find and classify the local minima, local maxima, and saddle points of the function f(x,y) = -3yx^2 - 3xy^2 + 36xy, we need to follow a step-by-step process:
Step 1: Calculate the first-order partial derivatives of the function with respect to x and y.
- Differentiating f(x,y) with respect to x, we get:
Fx = -6yx - 6y^2 + 36y
- Differentiating f(x,y) with respect to y, we get:
Fy = -6yx^2 - 6xy + 36x
Step 2: Set both Fx and Fy equal to zero and solve the resulting system of equations simultaneously.
Set Fx = 0:
-6yx - 6y^2 + 36y = 0
Set Fy = 0:
-6yx^2 - 6xy + 36x = 0
Step 3: Solve the system of equations to find critical points (points where both partial derivatives are zero).
Solving the system of equations is an algebraic task that involves substituting one equation into the other and solving for x and y values. Let's solve it step-by-step:
From Fx = 0:
-6yx - 6y^2 + 36y = 0
Factor out a -6y from the equation:
-6y(x + y - 6) = 0
This equation gives two possibilities:
1) -6y = 0, which implies y = 0
2) x + y - 6 = 0, which implies x = 6 - y
Substituting x = 6 - y into Fy = 0:
-6yx^2 - 6xy + 36x = 0
-6y(6 - y)^2 - 6(6 - y)y + 36(6 - y) = 0
Simplify and expand:
-6y(36 - 12y + y^2) - 6(6y - y^2) + 216 - 36y = 0
-216y + 72y^2 - 6y^3 - 36y + 6y^2 + 216 - 36y = 0
Combine like terms:
-6y^3 + 78y^2 - 288y + 216 = 0
Divide both sides by -6:
y^3 - 13y^2 + 48y - 36 = 0
Now we need to find the solutions to this cubic equation. There are various methods to solve cubic equations, such as factoring, synthetic division, or using numerical methods. The solutions to this equation will give you the y-values of the critical points.
Step 4: Substitute the found y-values back into the equation x = 6 - y to obtain the corresponding x-values of the critical points.
Substituting each y-value into x = 6 - y will give you the x-values of the critical points.
Step 5: Use the second-order partial derivatives test to classify the critical points as local minima, local maxima, or saddle points.
Calculate the second-order partial derivatives of f(x,y) with respect to x and y:
Fxx = -6y
Fyy = -6x
Fxy = -6x - 6y + 36
Evaluate the discriminant, D, using the second-order partial derivatives:
D = Fxx * Fyy - (Fxy)^2
D = (-6y) * (-6x) - (-6x - 6y + 36)^2
Simplify D.
- Analyze the signs of Fxx and D to classify the critical points:
- If Fxx > 0 and D > 0, it is a local minimum.
- If Fxx < 0 and D > 0, it is a local maximum.
- If D < 0, it is a saddle point.
By following these steps, you should be able to find and classify all local minima, local maxima, and saddle points for the given function f(x, y).