Final answer:
The function has a relative minimum at (-14, -28) and no relative maximum.
Step-by-step explanation:
To find the relative maximum and minimum values of the function f(x, y) = x² + xy + y² - 28y + 261, we need to find the critical points by taking the partial derivatives with respect to x and y and setting them equal to zero. Taking the partial derivative with respect to x, we get 2x + y = 0, and taking the partial derivative with respect to y, we get x + 2y - 28 = 0. Solving these equations simultaneously, we find x = -14, y = -28. To determine whether these are relative maximum or minimum points, we can use the second partial derivative test. Taking the second partial derivatives, we find that the second partial derivative with respect to x is 2 and the second partial derivative with respect to y is 2. Since both second partial derivatives are positive, the point (-14, -28) is a relative minimum.
To find the relative maximum, we can check the boundary of the region defined by the constraint. Since there is no constraint given, we can assume that x and y can take any value. However, since the function is a quadratic polynomial, it can be shown that there is no global maximum or minimum. Therefore, there is no relative maximum for this function.