Problem 1
1. Find and classify the critical points for \(f(x,y) = x^2+y^2-6x-14y+100\). (5 points)
Solution. We take the partial derivatives with respect to \(x\) and \(y\). We have \(f_x = 2x-6\) and \(f_y = 2y-14\), giving us a critical point of \((3,7)\). We now perform the second derivative test. Seeing that \(f_{xx}=2=f_{yy}\) and \(f_{xy}=0\), we get \(D=4\), so we have a relative minimum at \((3,7)\).
Problem 2
2. Find the absolute maximum and minimum values for \(f(x,y) = \dfrac{-2y}{x^2+y^2+1}\) on \(D = \{(x,y)\in\mathbb{R}^2\ |\ x^2+y^2\le 4\}\). (5 points)
Solution. We begin by finding the critical points within the boundary. The partial derivatives are
\[ f_x = \frac{4xy}{(x^2+y^2+1)^2} \quad \text{and} \quad f_y = \frac{-2(x^2-y^2+1)}{(x^2+y^2+1)^2}. \]
To find out when these give critical points, note that the denominators will never be zero. Thus, we need to only find the zeros of the above numerators. In particular, \(f_x=0\) only along the \(x\) and \(y\) axes. Note that if \(y=0\), then the numerator of \(f_y\) is \(-2(x^2+1)\), which cannot be zero. Thus, we restrict our attention to the case that \(x=0\), where now the numerator of \(f_y\) is \(-2(1-y^2)\) which has zeros at \(y^2=1\), meaning our critical points are \((0,1)\) and \((0,-1)\). As for the boundary, note that along \(x^2+y^2=4\), \(f(x,y)\) is of the form \(f(x,y) = -2y/5\). This is a linear function in \(y\), with constant derivative, so the absolute extrema will lie on the boundary, that is, \(y=\pm 2\). Thus, our 4 points to check are \((0,1)\), \((0,-1)\), \((0,2)\), \((0,-2)\). We see that the first two points, producing \(-1\) and \(1\) respectively, are the absolute extrema, as the points on the boundary only give \(-4/5\) and \(4/5\) respectively.