find the absolute minimum and absolute maximum of the function ( f(x,y)=xy - 3y - 9x + 27 ) on the region on…

Explanation:

Step1: Find the critical points in the interior

First, find the partial derivatives of (f(x,y)=xy - 3y-9x + 27). The partial derivative with respect to (x) is (f_x=y - 9), and the partial derivative with respect to (y) is (f_y=x - 3). Set (f_x = 0) and (f_y=0), so (y - 9=0) gives (y = 9), and (x - 3=0) gives (x = 3). The critical point is ((3,9)), but (y=x^{2}), when (x = 3), (y=9) (on the boundary (y=x^{2})).

Step2: Parameterize the boundaries

Boundary 1: (y=x^{2})

Substitute (y=x^{2}) into (f(x,y)), we get (g(x)=x\cdot x^{2}-3x^{2}-9x + 27=x^{3}-3x^{2}-9x + 27). Take the derivative (g^{\prime}(x)=3x^{2}-6x - 9=3(x^{2}-2x - 3)=3(x - 3)(x + 1)). Set (g^{\prime}(x)=0), then (x=3) or (x=-1). When (x = 3), (y = 9); when (x=-1), (y = 1). (g(3)=3^{3}-3\times3^{2}-9\times3 + 27=0), (g(-1)=(-1)^{3}-3\times(-1)^{2}-9\times(-1)+27=-1 - 3 + 9+27=32).

Boundary 2: (y = 14)

Substitute (y = 14) into (f(x,y)), we get (h(x)=14x-3\times14-9x + 27=5x - 15). Since (y=x^{2}\leq14), then (x\in[-\sqrt{14},\sqrt{14}]). (h(x)) is a linear function. (h(-\sqrt{14})=-5\sqrt{14}-15\approx-5\times3.74 - 15=-18.7-15=-33.7), (h(\sqrt{14})=5\sqrt{14}-15\approx5\times3.74-15 = 18.7-15 = 3.7).

Step3: Compare the function values

We have (f(3,9)=0), (f(-1,1)=(-1)\times1-3\times1-9\times(-1)+27=-1-3 + 9+27=32), (f(-\sqrt{14},14)\approx-33.7), (f(\sqrt{14},14)\approx3.7)

Answer:

Absolute minimum value: (-33.7) (attained at ((-\sqrt{14},14))), Absolute maximum value: (32) (attained at ((-1,1)))