Find all the local maxima, local minima, and saddle points of the function. \[ f(x, y)=x^{2}+x y+y^{2}+4 x-7 y+7 \] Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. A local maximum occurs at (Type an ordered pair. Use a comma to separate answers as needed.) The local maximum value(s) is/are (Type an exact answer. Use a comma to separate answers as needed.) B. There are no local maxima.

Step 1 ：First, we need to find the critical points of the function. The critical points are where the first derivatives of the function are equal to zero. So we take the partial derivatives of the function with respect to \(x\) and \(y\) and set them equal to zero.

Step 2 ：\[\frac{\partial f}{\partial x} = 2x + y + 4 = 0\]

Step 3 ：\[\frac{\partial f}{\partial y} = x + 2y - 7 = 0\]

Step 4 ：Solving these two equations simultaneously, we get \(x = -1\) and \(y = 3\). So the critical point is \((-1, 3)\).

Step 5 ：Next, we need to determine whether this critical point is a local maximum, local minimum, or saddle point. We do this by finding the second derivatives of the function and using them to compute the determinant of the Hessian matrix.

Step 6 ：The second derivatives are:

Step 7 ：\[\frac{\partial^2 f}{\partial x^2} = 2\]

Step 8 ：\[\frac{\partial^2 f}{\partial y^2} = 2\]

Step 9 ：\[\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 1\]

Step 10 ：The Hessian matrix is then:

Step 11 ：\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\]

Step 12 ：The determinant of the Hessian matrix is \(\det(H) = (2)(2) - (1)(1) = 3\).

Step 13 ：Since \(\det(H) > 0\) and \(\frac{\partial^2 f}{\partial x^2} > 0\), the critical point \((-1, 3)\) is a local minimum.

Step 14 ：Finally, we substitute \(x = -1\) and \(y = 3\) into the function to find the local minimum value.

Step 15 ：\[f(-1, 3) = (-1)^2 + (-1)(3) + 3^2 + 4(-1) - 7(3) + 7 = -6\]

Step 16 ：So, the local minimum occurs at \((-1, 3)\) and the local minimum value is \(-6\).

Step 17 ：Since there is only one critical point and it is a local minimum, there are no local maxima or saddle points. So, the answer is \(\boxed{\text{B. There are no local maxima.}}\)