Question

find the absolute minimum and absolute maximum of the function ( f(x,y)=xy - 3y - 9x + 27 ) on the region on or above ( y = x^{2} ) and on or below ( y = 14 ) and list the points where they occur. if the absolute min or max is attained at multiple points list them all, separated by commas. absolute minimum value: attained at. absolute maximum value: attained at.

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 \((-1,1)\))