To find the local maximum, minimum, and saddle points of the function $f(x, y) = x^3y + 12x^2 - 8y$, we proceed as follows:

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Step 1: Compute the First Partial Derivatives

The first-order partial derivatives are: $f_x = \frac{\partial f}{\partial x} = 3x^2y + 24x, \quad f_y = \frac{\partial f}{\partial y} = x^3 - 8.$

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Step 2: Solve the Critical Points

Critical points occur where both partial derivatives are zero:

1. $f_x = 3x^2y + 24x = 0$,
2. $f_y = x^3 - 8 = 0$.

Solving $f_y = 0$:

$x^3 - 8 = 0 \implies x = 2.$

Substituting $x = 2$ into $f_x = 0$:

$3(2)^2y + 24(2) = 0 \implies 12y + 48 = 0 \implies y = -4.$

Thus, the only critical point is: $(x, y) = (2, -4).$

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Step 3: Compute the Second Partial Derivatives

The second-order partial derivatives are: $f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6xy + 24, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2} = 0, \quad f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 3x^2.$

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Step 4: Analyze Using the Hessian Determinant

The Hessian determinant is given by: $H = f_{xx}f_{yy} - (f_{xy})^2.$

Substitute the critical point $(2, -4)$:

• $f_{xx} = 6(2)(-4) + 24 = -48 + 24 = -24$,
• $f_{yy} = 0$,
• $f_{xy} = 3(2)^2 = 12$.

Thus, $H = f_{xx}f_{yy} - (f_{xy})^2 = (-24)(0) - (12)^2 = -144.$

Since $H < 0$, the critical point $(2, -4)$ is a saddle point.

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Final Answer:

• The function $f(x, y)$ has no local maximum or minimum.
• There is a saddle point at $(2, -4)$.

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Would you like me to provide a detailed explanation of the Hessian determinant or any additional steps?

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Follow-Up Questions:

1. What is the significance of the Hessian determinant in classifying critical points?
2. How can the method be extended to three-variable functions?
3. What happens if the Hessian determinant is zero at a critical point?
4. Can you graph $f(x, y)$ to visualize the saddle point?
5. What is the impact of modifying the coefficients in $f(x, y)$ on the critical points?

Tip:

When the Hessian determinant is negative, the function has a saddle point at that critical location.