91Ó°ÊÓ

A partial derivative is a derivative taken of a function with multiple variables, with respect to one variable while treating the others as constants. It helps quantify how a function changes as individual variables change. In the given function, \( f(x, y) = x^4 + y^4 - 4x - 32y + 10 \), we compute the partial derivatives with respect to \( x \) and \( y \) to find the critical points. This process is crucial because critical points are where the function potentially reaches local maxima, minima, or saddle points.

For our function, the partial derivatives are:
\( f_x = 4x^3 - 4 \),
\( f_y = 4y^3 - 32 \).

Setting these equations equal to zero and solving them helps us locate critical points where the function's slope is zero. In this case, the partial derivative with respect to \( x \) gives \( x^3 = 1 \), leading to \( x = 1 \), and for \( y \), the equation \( y^3 = 8 \) results in \( y = 2 \). Thus, the critical point is at \( (1, 2) \).

The Second Derivative Test is a valuable tool for classifying critical points found from partial derivatives. It helps determine if a critical point is a local maximum, a local minimum, or a saddle point. This test uses second partial derivatives to evaluate the nature of a critical point.

The test computes a determinant, \( D \), from the second partial derivatives:
\[ D = f_{xx}(a,b) \cdot f_{yy}(a,b) - (f_{xy}(a,b))^2 \]
where:

• \( f_{xx} \) and \( f_{yy} \) are the second partial derivatives with respect to \( x \) and \( y \), respectively.
• \( f_{xy} \) and \( f_{yx} \) are mixed partial derivatives.

For our example, the second partial derivatives are:
\( f_{xx} = 12x^2 \),
\( f_{yy} = 12y^2 \),
\( f_{xy} = f_{yx} = 0 \).
At the critical point \( (1, 2) \), the test value \( D = 12 \times 1^2 \times 12 \times 2^2 - 0^2 = 144 \).

If \( D > 0 \) and \( f_{xx} > 0 \), the critical point is a local minimum. If \( D > 0 \) and \( f_{xx} < 0 \), it is a local maximum. If \( D < 0 \), the point is a saddle point. If \( D = 0 \), the test is inconclusive.

A local minimum refers to a point in the domain of a function where the function value is less than or equal to nearby points, indicating a trough or dip. At this point, the function temporarily decreases and subsequently increases.

In the exercise above, after evaluating the Second Derivative Test, we found that at \( (1, 2) \), we have a local minimum.

• The determinant \( D = 144 \) is greater than zero, confirming concavity.
• The second derivative \( f_{xx}(1, 2) = 12 \) is also greater than zero, signifying the region around the critical point curves upward.

These results indicate the presence of a local minimum at \( (1, 2) \) in the function \( f(x, y) = x^4 + y^4 - 4x - 32y + 10 \). Here, the value of the function is at its local lowest compared to its immediate surrounding values.