To find the local extreme values of the function f(x, y) = x^3 + y^3 - 3x - 12y + 20, we need to follow these steps:

Step 1: Find the partial derivatives of the function with respect to x and y. The partial derivative with respect to x (denoted as ∂f/∂x) is obtained by differentiating the function with respect to x while treating y as a constant: ∂f/∂x = 3x^2 - 3.

The partial derivative with respect to y (denoted as ∂f/∂y) is obtained by differentiating the function with respect to y while treating x as a constant: ∂f/∂y = 3y^2 - 12.

Step 2: Set the partial derivatives equal to zero and solve for x and y. Setting ∂f/∂x = 0, we have: 3x^2 - 3 = 0. Solving this equation, we find: x^2 = 1, which gives us two possible values for x: x = 1 and x = -1.

Setting ∂f/∂y = 0, we have: 3y^2 - 12 = 0. Solving this equation, we find: y^2 = 4, which gives us two possible values for y: y = 2 and y = -2.

Step 3: Calculate the second partial derivatives. The second partial derivative with respect to x (denoted as ∂^2f/∂x^2) is obtained by differentiating ∂f/∂x with respect to x: ∂^2f/∂x^2 = 6x.

The second partial derivative with respect to y (denoted as ∂^2f/∂y^2) is obtained by differentiating ∂f/∂y with respect to y: ∂^2f/∂y^2 = 6y.

Step 4: Evaluate the second partial derivatives at the critical points. For x = 1, we have ∂^2f/∂x^2 = 6(1) = 6. For x = -1, we have ∂^2f/∂x^2 = 6(-1) = -6. For y = 2, we have ∂^2f/∂y^2 = 6(2) = 12. For y = -2, we have ∂^2f/∂y^2 = 6(-2) = -12.

Step 5: Determine the nature of the critical points. To determine the nature of the critical points, we use the second partial derivative test:

If ∂^2f/∂x^2 > 0 and ∂^2f/∂y^2 > 0, then the critical point is a local minimum.

If ∂^2f/∂x^2 < 0 and ∂^2f/∂y^2 < 0, then the critical point is a local maximum.

If ∂^2f/∂x^2 and ∂^2f/∂y^2 have different signs, then the critical point is a saddle point.

For x = 1 and y = 2, we have ∂^2f/∂x^2 = 6 > 0 and ∂^2f/∂y^2 = 12 > 0, so this critical point is a local minimum.

For x = 1 and y = -2, we have ∂^2f/∂x^2 = 6 > 0 and ∂^2f/∂y^2 = -12 < 0, so this critical point is a saddle point.

For x = -1 and y = 2, we have ∂^2f/∂x^2 = -6 < 0 and ∂^2f/∂y^2 = 12 > 0, so this critical point is a saddle point.

For x = -1 and y = -2, we have ∂^2f/∂x^2 = -6 < 0 and ∂^2f/∂y^2 = -12 < 0, so this critical point is a local maximum.

Therefore, the local extreme values of the function f(x, y) = x^3 + y^3 - 3x - 12y + 20 are:

Local minimum at (1, 2).

Saddle points at (1, -2) and (-1, 2).

Local maximum at (-1, -2).

Question

To find the local extreme values of the function f(x, y) = x^3 + y^3 - 3x - 12y + 20, we need to follow these steps:

Step 1: Find the partial derivatives of the function with respect to x and y. The partial derivative with respect to x (denoted as ∂f/∂x) is obtained by differentiating the function with respect to x while treating y as a constant: ∂f/∂x = 3x^2 - 3.

The partial derivative with respect to y (denoted as ∂f/∂y) is obtained by differentiating the function with respect to y while treating x as a constant: ∂f/∂y = 3y^2 - 12.

Step 2: Set the partial derivatives equal to zero and solve for x and y. Setting ∂f/∂x = 0, we have: 3x^2 - 3 = 0. Solving this equation, we find: x^2 = 1, which gives us two possible values for x: x = 1 and x = -1.

Setting ∂f/∂y = 0, we have: 3y^2 - 12 = 0. Solving this equation, we find: y^2 = 4, which gives us two possible values for y: y = 2 and y = -2.

Step 3: Calculate the second partial derivatives. The second partial derivative with respect to x (denoted as ∂^2f/∂x^2) is obtained by differentiating ∂f/∂x with respect to x: ∂^2f/∂x^2 = 6x.

The second partial derivative with respect to y (denoted as ∂^2f/∂y^2) is obtained by differentiating ∂f/∂y with respect to y: ∂^2f/∂y^2 = 6y.

Step 4: Evaluate the second partial derivatives at the critical points. For x = 1, we have ∂^2f/∂x^2 = 6(1) = 6. For x = -1, we have ∂^2f/∂x^2 = 6(-1) = -6. For y = 2, we have ∂^2f/∂y^2 = 6(2) = 12. For y = -2, we have ∂^2f/∂y^2 = 6(-2) = -12.

Step 5: Determine the nature of the critical points. To determine the nature of the critical points, we use the second partial derivative test:

If ∂^2f/∂x^2 > 0 and ∂^2f/∂y^2 > 0, then the critical point is a local minimum.
If ∂^2f/∂x^2 < 0 and ∂^2f/∂y^2 < 0, then the critical point is a local maximum.
If ∂^2f/∂x^2 and ∂^2f/∂y^2 have different signs, then the critical point is a saddle point.

For x = 1 and y = 2, we have ∂^2f/∂x^2 = 6 > 0 and ∂^2f/∂y^2 = 12 > 0, so this critical point is a local minimum.

For x = 1 and y = -2, we have ∂^2f/∂x^2 = 6 > 0 and ∂^2f/∂y^2 = -12 < 0, so this critical point is a saddle point.

For x = -1 and y = 2, we have ∂^2f/∂x^2 = -6 < 0 and ∂^2f/∂y^2 = 12 > 0, so this critical point is a saddle point.

For x = -1 and y = -2, we have ∂^2f/∂x^2 = -6 < 0 and ∂^2f/∂y^2 = -12 < 0, so this critical point is a local maximum.

Therefore, the local extreme values of the function f(x, y) = x^3 + y^3 - 3x - 12y + 20 are:

Local minimum at (1, 2).
Saddle points at (1, -2) and (-1, 2).
Local maximum at (-1, -2).

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