Find the absolute maxima and minima of the function on the given domain.
-$$f ( x , y ) = x ^ { 2 } + 12 x + y ^ { 2 } + 16 y + 10 \text { on the rectangular region } - 1 \leq x \leq 1 , - 2 \leq y \leq 2$$
A) Absolute maximum: 74 at (2, 2); absolute minimum: 10 at (0, 0)
B) Absolute maximum: 74 at (2, 2); absolute minimum: -29 at (-1, -2)
C) Absolute maximum: 59 at (1, 2); absolute minimum: -29 at (-1, -2)
D) Absolute maximum: 59 at (1, 2); absolute minimum: 10 at (0, 0)

Question

Quizplus · Accepted Answer

To find the absolute maxima and minima of the function $f(x, y) = x^2 + 12x + y^2 + 16y + 10$ on the given domain, we first complete the square for both $x$ and $y$. The function can be rewritten as $f(x, y) = (x^2 + 12x + 36) + (y^2 + 16y + 64) - 90 = (x + 6)^2 + (y + 8)^2 - 90$. The minimum value occurs at the vertex of the parabola, which is at $(-6, -8)$, but this point is outside the given domain. Therefore, we check the boundaries of the domain. At $x = -1$, $f(x, y) = (-1)^2 + 12(-1) + y^2 + 16y + 10 = y^2 + 16y - 1$, which has its minimum at $y = -8$ (outside the domain) and needs to be evaluated at $y = -2$ and $y = 2$. Similarly, we evaluate the function at $x = 1$, $y = -2$, and $y = 2$. The maximum value within the given domain occurs at the corners because the function increases as we move away from $(-6, -8)$. Evaluating the function at the corners of the domain, we find that the maximum occurs at $x = 1, y = 2$ and the minimum occurs at $x = -1, y = -2$. Evaluating $f(1, 2) = 1^2 + 12(1) + 2^2 + 16(2) + 10 = 1 + 12 + 4 + 32 + 10 = 59$ for the maximum, and $f(-1, -2) = (-1)^2 + 12(-1) + (-2)^2 + 16(-2) + 10 = 1 - 12 + 4 - 32 + 10 = -29$ for the minimum. Therefore, the correct answer is the absolute maximum of 59 at (1, 2) and the absolute minimum of -29 at (-1, -2).

Find the Absolute Maxima and Minima of the Function on the Given