Find and classify all local minima, local maxima, and saddle points of the function f(x,y)= -3yx^2-3xy^2+36xy
F = -3yx^2-3xy^2+36xy
Fx = -6xy - 3y^2 + 36y
Fxx = -6y
Fy = -3x^2 - 6xy + 36x
Fyy = -6x
Fxy = -6x - 6y + 36
D = FxxFyy-(Fxy)^2 = 36xy - 36(x+y-6)^2
Fx = 0 Fy=0 at (0,0)
D<0 so a saddle point
Fx = -6xy - 3y^2 + 36y
Fx = 0 when y = 2(6-x)
Fy = -3x^2 - 6xy + 36x
Fy = 0 when x = 2(6-y)
So there is a local max for z along those two lines
Okay Im just confused as how to get Fx=0 and Fy=0, If I set -6xy-3y^2+36y=0 how do i solve this?? Algebra was so long ago!
could you detail that part of the problem...
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 calculate the partial derivatives and set them equal to zero to find the critical points. Then, we can use the second derivative test to classify these points.
Step 1: Calculate the first-order partial derivatives:
∂f/∂x = -6xy - 3y^2 + 36y
∂f/∂y = -3x^2 - 6xy + 36x
Step 2: Set the partial derivatives equal to zero and solve for x and y:
-6xy - 3y^2 + 36y = 0 -> 2xy + y(12 - y) - 12y = 0 -> 2xy - y^2 - 12y = 0 -> y(2x - y - 12) = 0
-3x^2 - 6xy + 36x = 0 -> -x^2 - 2xy + 12x = 0 -> -x(x + 2y - 12) = 0
Solving these equations, we find the following critical points:
1. y = 0 or 2x - y - 12 = 0
2. x = 0 or x + 2y - 12 = 0
Step 3: Classify the critical points using the second derivative test.
To determine the classification, we need to calculate the second-order partial derivatives:
∂^2f/∂x^2 = -6y
∂^2f/∂y^2 = -6x
∂^2f/∂x∂y = -6x - 6y + 36
Step 4: Evaluate the second-order partial derivatives at each critical point.
For x = 0, y = 0:
∂^2f/∂x^2 = -6(0) = 0
∂^2f/∂y^2 = -6(0) = 0
∂^2f/∂x∂y = -6(0) - 6(0) + 36 = 36
For x = 0, x + 2y - 12 = 0:
∂^2f/∂x^2 = -6(2y - 12) = -12y + 72
∂^2f/∂y^2 = -6(0) = 0
∂^2f/∂x∂y = -6(0) - 6(2y - 12) + 36 = -12y + 84
For y = 0, 2x - y - 12 = 0:
∂^2f/∂x^2 = -6(0) = 0
∂^2f/∂y^2 = -6(2x - 12) = -12x + 72
∂^2f/∂x∂y = -6(0) - 6(0) + 36 = 36
For x + 2y - 12 = 0, 2x - y - 12 = 0:
∂^2f/∂x^2 = -6(y) = -6y
∂^2f/∂y^2 = -6(x) = -6x
∂^2f/∂x∂y = -6(x) - 6(y) + 36 = -6x - 6y + 36
Step 5: Classify the critical points.
Case 1: For y = 0 and x = 0:
The classification cannot be determined by the second derivative test as the second partial derivatives are both zero.
Case 2: For y = 0 and 2x - y - 12 = 0:
∂^2f/∂x^2 = -12y + 72 = 72
∂^2f/∂y^2 = -12x + 72 = 72
∂^2f/∂x∂y = -12y + 84 = 84
The second derivative test tells us that this critical point is a local minimum since ∂^2f/∂x^2 > 0 and (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 > 0.
Case 3: For x = 0 and x + 2y - 12 = 0:
∂^2f/∂x^2 = -12y + 72 = 72
∂^2f/∂y^2 = -6x = 0
∂^2f/∂x∂y = -6y + 36 = 36
The second derivative test tells us that this critical point is a saddle point since (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 < 0.
Case 4: For x + 2y - 12 = 0 and 2x - y - 12 = 0:
∂^2f/∂x^2 = -6y = 0
∂^2f/∂y^2 = -6x = 0
∂^2f/∂x∂y = -6x - 6y + 36 = 36
The second derivative test cannot determine the nature of this critical point since both (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 and ∂^2f/∂x^2 are zero.
To summarize:
- The point (0, 0) could be a local minimum or a saddle point. Further analysis is required.
- The point (0, 6) is a local minimum.
- The point (12, 0) is a saddle point.
- The point (6, 6) cannot be determined using the second derivative test and requires further analysis.
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 find the critical points and then analyze the second-order partial derivatives at those points.
Step 1: Find the first-order partial derivatives:
∂f/∂x = -6xy - 3y^2 + 36y
∂f/∂y = -3x^2 - 6xy + 36x
Step 2: Find the critical points:
To find the critical points, set both partial derivatives equal to zero:
-6xy - 3y^2 + 36y = 0 .............(1)
-3x^2 - 6xy + 36x = 0 .............(2)
Solving equation (1) for y, we get:
-6xy - 3y^2 + 36y = 0
y(-6x - 3y + 36) = 0
y = 0 or -6x - 3y + 36 = 0
If y = 0, substitute this into equation (2) and solve for x:
-3x^2 - 6xy + 36x = 0
-3x^2 + 36x = 0
-3x(x - 12) = 0
x = 0 or x = 12
If -6x - 3y + 36 = 0, substitute y = -6x + 36 into equation (2) and solve for x:
-3x^2 - 6xy + 36x = 0
-3x^2 - 6x(-6x + 36) + 36x = 0
-3x^2 + 36x^2 - 216x + 36x = 0
33x^2 - 180x = 0
3x(11x - 60) = 0
x = 0 or x = 60/11
Substitute the values of x back into the equation -6xy - 3y^2 + 36y = 0 to find the corresponding y-values:
For x = 0:
-6(0)y - 3y^2 + 36y = 0
-3y^2 + 36y = 0
3y(y - 12) = 0
y = 0 or y = 12
For x = 12:
-6(12)y - 3y^2 + 36y = 0
-72y - 3y^2 + 36y = 0
-3y^2 + 36y - 72y = 0
-3y^2 - 36y = 0
-3y(y + 12) = 0
y = 0 or y = -12
For x = 60/11:
-6(60/11)y - 3y^2 + 36y = 0
-360y - 3y^2 + 36y = 0
-3y^2 - 324y = 0
-3y(y + 108) = 0
y = 0 or y = -108
Step 3: Analyze the second-order partial derivatives at the critical points:
To classify the critical points, we need to analyze the second-order partial derivatives using the second-order partial derivatives test. The second-order partial derivatives are as follows:
∂^2f/∂x^2 = -6y
∂^2f/∂y^2 = -6x
∂^2f/∂x∂y = -6x - 6y + 36
For each critical point, substitute the x and y values into the second-order partial derivatives.
For (x, y) = (0, 0):
∂^2f/∂x^2 = -6(0) = 0
∂^2f/∂y^2 = -6(0) = 0
∂^2f/∂x∂y = -6(0) - 6(0) + 36 = 36
Since the discriminant (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = 0(0) - 36^2 < 0, the point (0,0) is a saddle point.
For (x, y) = (0, 12):
∂^2f/∂x^2 = -6(12) = -72
∂^2f/∂y^2 = -6(0) = 0
∂^2f/∂x∂y = -6(0) - 6(12) + 36 = -72
Since the discriminant (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = (-72)(0) - (-72)^2 > 0 and ∂^2f/∂x^2 = -72 < 0, the point (0,12) is a local maximum.
For (x, y) = (12, 0):
∂^2f/∂x^2 = -6(0) = 0
∂^2f/∂y^2 = -6(12) = -72
∂^2f/∂x∂y = -6(12) - 6(0) + 36 = -72
Since the discriminant (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = (0)(-72) - (-72)^2 > 0 and ∂^2f/∂y^2 = -72 < 0, the point (12,0) is a local maximum.
For (x, y) = (60/11, 0):
∂^2f/∂x^2 = -6(0) = 0
∂^2f/∂y^2 = -6(60/11) = -360/11
∂^2f/∂x∂y = -6(60/11) - 6(0) + 36 = -360/11
Since the discriminant (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = (0)(-360/11) - (-360/11)^2 < 0, the point (60/11,0) is a saddle point.
For (x, y) = (0, -12):
∂^2f/∂x^2 = -6(-12) = 72
∂^2f/∂y^2 = -6(0) = 0
∂^2f/∂x∂y = -6(0) - 6(-12) + 36 = 72
Since the discriminant (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = (72)(0) - 72^2 > 0 and ∂^2f/∂x^2 = 72 > 0, the point (0,-12) is a local minimum.
For (x, y) = (60/11, -108):
∂^2f/∂x^2 = -6(-108) = 648
∂^2f/∂y^2 = -6(60/11) = -360/11
∂^2f/∂x∂y = -6(60/11) - 6(-108) + 36 = 648/11
Since the discriminant (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = (648)(-360/11) - (648/11)^2 < 0, the point (60/11,-108) is a saddle point.
In summary:
The function f(x, y) = -3yx^2 - 3xy^2 + 36xy has the following critical points:
- Local minimum: (0, -12)
- Local maximum: (0, 12), (12, 0)
- Saddle points: (0, 0), (60/11, 0), (60/11, -108)