Introduction to Local Maxima and Minima in Multivariate Functions

In the realm of mathematical optimization and calculus, understanding the local maxima and minima of a function is fundamental. This article delves into a detailed analysis of the function z x^2y - xy^2 - 4xy - 4x^2 - 4y^2 and how to identify its critical points, specifically focusing on the local maxima and local minima.

Step 1: Derivation of Partial Derivatives

To find the critical points, we start by computing the first partial derivatives of the function with respect to x and y. The function in question is:

z x^2y - xy^2 - 4xy - 4x^2 - 4y^2

First partial derivative with respect to x is:

(frac{partial z}{partial x} 2xy - y^2 - 4y - 8x)

And the first partial derivative with respect to y is:

(frac{partial z}{partial y} x^2 - 2xy - 4x - 8y)

Step 2: Setting the Derivatives to Zero

To locate the critical points, we set both partial derivatives equal to zero and solve the resulting system of equations:

(2xy - y^2 - 4y - 8x 0) (x^2 - 2xy - 4x - 8y 0)

These equations can be complex to solve directly. Instead, we explore simpler solutions by inspection.

Step 3: Simplifying the System of Equations

Lets try y 0:

(0 8x Rightarrow x 0)

This gives us one critical point: (0, 0).

Lets try x 0:

(-8y 0 Rightarrow y 0)

This confirms (0, 0) again. Additional critical points can be found through numerical methods or further inspection.

Step 4: Descending into the Second Derivative Test

To classify the critical points, we utilize the second derivative test. We first need the second partial derivatives:

Second partial derivative with respect to x: (frac{partial^2 z}{partial x^2} 2y - 8) Second partial derivative with respect to y: (frac{partial^2 z}{partial y^2} -2x - 8) Second mixed partial derivative: (frac{partial^2 z}{partial x partial y} 2x - 2y - 4)

Evaluating these at the critical point (0, 0) yields:

(frac{partial^2 z}{partial x^2}0, 0 2(0) - 8 -8) (frac{partial^2 z}{partial y^2}0, 0 -2(0) - 8 -8) (frac{partial^2 z}{partial x partial y}0, 0 2(0) - 2(0) - 4 -4)

Step 5: Calculating the Determinant of the Hessian Matrix

The determinant (D) of the Hessian matrix is:

(D frac{partial^2 z}{partial x^2} cdot frac{partial^2 z}{partial y^2} - left(frac{partial^2 z}{partial x partial y}right)^2)

Evaluating (D) at (0, 0) gives:

D (-8)(-8) - (-4)^2 64 - 16 48)

Since (D > 0) and (frac{partial^2 z}{partial x^2}

Conclusion

In conclusion, the function (z x^2y - xy^2 - 4xy - 4x^2 - 4y^2) has a local maximum at the point (0, 0). Further analysis may be required to locate additional critical points, but this point is confirmed to be a local maximum. This analysis provides a comprehensive approach to identifying local maxima and minima in multivariate functions.