Find the critical points of the given function and classify each as a relative maximum, a relative minimum, or a saddle point. If the function has extremes, calculate max and/or…

To find the critical points and classify them for the given function \( f(x, y) = -4x^2 + 12xy - y + 48x + 6y - 7 \), we will follow these steps:

1. Find the first partial derivatives of the function with respect to \( x \) and \( y \).
2. Set the first partial derivatives to zero to find the critical points.
3. Find the second partial derivatives to classify the critical points using the second derivative test.

Step 1: Calculate the first partial derivatives

The function is given by: \[ f(x, y) = -4x^2 + 12xy - y + 48x + 6y - 7 \]

Calculating the first partial derivatives:

\[ f_x = \frac{\partial f}{\partial x} = -8x + 12y + 48 \]

\[ f_y = \frac{\partial f}{\partial y} = 12x - 1 + 6 = 12x + 5 \]

Step 2: Set the first partial derivatives to zero

Set \( f_x \) and \( f_y \) to zero:

1. \( -8x + 12y + 48 = 0 \)
2. \( 12x + 5 = 0 \)

From the second equation, we can solve for \( x \): \[ 12x + 5 = 0 \implies 12x = -5 \implies x = -\frac{5}{12} \]

Now, substitute \( x = -\frac{5}{12} \) into the first equation: \[ -8\left(-\frac{5}{12}\right) + 12y + 48 = 0 \] \[ \frac{40}{12} + 12y + 48 = 0 \] \[ \frac{10}{3} + 12y + 48 = 0 \] \[ 12y = -\frac{10}{3} - 48 \] Convert \( 48 \) to a fraction: \[ 48 = \frac{144}{3} \implies 12y = -\frac{10}{3} - \frac{144}{3} = -\frac{154}{3} \] \[ y = -\frac{154}{36} = -\frac{77}{18} \]

The critical point is: \[ \left(-\frac{5}{12}, -\frac{77}{18}\right) \]

Step 3: Classify the critical point using the second derivative test

Calculate the second partial derivatives: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = -8 \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 0 \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 12 \]

Now, compute the determinant of the Hessian matrix \( D \): \[ D = f_{xx} f_{yy} - (f_{xy})^2 = (-8)(0) - (12)^2 = -144 \]

Since \( D < 0 \), this indicates that the critical point \( \left(-\frac{5}{12}, -\frac{77}{18}\right) \) is a saddle point.

Conclusion

• The critical point is \( \left(-\frac{5}{12}, -\frac{77}{18}\right) \).
• It is classified as a saddle point.
• The function does not have relative maximum or minimum values.