91Ó°ÊÓ

To find possible relative maxima or minima, we first need to determine the partial derivatives of the function with respect to each variable. In our given function:

\[f(x, y)= 4 x^{2} + 4 x y - 3 y^{2} + 4 y - 1\]A partial derivative is like taking the derivative, but we treat all other variables as constants. The partial derivative with respect to \(x\) is:

\[\frac{\text{∂f}}{\text{∂x}}= 8x + 4y\]And, the partial derivative with respect to \(y\) is:

\[\frac{\text{∂f}}{\text{∂y}} = 4x - 6y + 4\]Partial derivatives help us understand how the function changes as we slightly move in the direction of one of the axes while keeping the other variable fixed. This is crucial for finding critical points where the function might reach its maximum or minimum.

Once we have the partial derivatives, the next step is to set them equal to zero to find the critical points. In other words, we solve the system of equations where:

• \[8x + 4y = 0\]
• \[4x - 6y + 4 = 0\]

Setting the partial derivatives to zero is necessary because it helps find the points where the function's rate of change with respect to each variable is zero. These points are where the function could potentially have a maximum or minimum.

We solve this system by expressing one variable in terms of the other. From the first equation:

\[y = -2x\]Next, we substitute \(y = -2x\) into the second equation:

\[4x - 6(-2x) + 4 = 0\]
\[4x + 12x + 4 = 0\]
\[16x + 4 = 0\]
\[16x = -4\]
\[x = -\frac{1}{4}\]

Then, substitute \(x = -\frac{1}{4}\) back into \(y = -2x\):

\[y = -2\bigg(-\frac{1}{4}\bigg) = \frac{1}{2}\]

This gives us the critical point \(\bigg(-\frac{1}{4}, \frac{1}{2}\bigg)\).

After finding the critical points, we need to determine whether each point is a relative maximum, minimum, or a saddle point. This requires evaluating the second partial derivatives: \(\frac{∂^2f}{∂x^2}\), \(\frac{∂^2f}{∂y^2}\), and \(\frac{∂^2f}{∂x∂y}\).

Using the second derivative test, if

\[D = f_{xx}(a, b)f_{yy}(a, b) - (f_{xy}(a, b))^2\]

where:

• \[D > 0\] and \[f_{xx}(a, b) > 0\], the point \((a, b)\) is a relative minimum.


• \[D > 0\] and \[f_{xx}(a, b) < 0\], the point \((a, b)\) is a relative maximum.


• \[D < 0\], the point \((a, b)\) is a saddle point (neither a maximum nor a minimum).

In our problem, you'd compute \(f_{xx}(x, y) = 8\), \(f_{yy}(x, y) = -6\), and \(f_{xy}(x, y) = 4\). Plug these back into the formula to determine the nature of the critical point \(\bigg(-\frac{1}{4}, \frac{1}{2}\bigg)\). With this information, you can classify each critical point and understand the shape and behavior of the original function .