Solution

The region is a circle of radius 2 centered at the origin. By the Extreme Value Theorem, there exists an absolute maximum and minumum since the circle and its interior is a closed region.

We check for critical points inside the interior of the circle using the Second Derivative Test to find relative maxima and minima. The calculation of partial derivatives give:

\[\begin{align} f_x & = 2x + 2 & \text{ } f_y & = 2y + 2 \\ f_{xx} & = 2 & \text{ } f_{yy} & = 2 \\ f_{xy} & = 0 & & \\ \end{align} \] To find critical points of the function on the boundary circle, we use the Lagrange Multiplier method.

The discriminant \[D = f_{xx}f_{yy} - f_{xy}^2 = 4 - 0 = 4 > 0 \] is always defined and positive and the 2nd partial derivative \(f_{xx} = 2 > 0\), so any critical point \(P\) must be a relative minimum.

The conditions \(f_x = 0\) and \(f_y = 0\) imply

\[\begin{align} 2x + 2 & = 0 & \implies & x = -1 \\ 2y + 2 & = 0 & \implies & y = -1 \\ \end{align} \]

At the point \(P=(-1,-1)\) we see that \(f(P) = f(-1,-1) = -2\). Thus, there is a relative minimum -2 at (-1,-1).

Next we check for critical points on the boundary curve using the method of Lagrange multipliers. Let \(g(x,y)\) denote the boundary condition.

\[ g(x,y) = 0 \implies g(x,y) = x^2 + y^2 -4 = 0 \] The Lagrange multiplier gives the equations: \[\begin{align} \nabla f(x,y) & = \lambda \nabla g(x,y) \\ g(x,y) & = 0 \\ \end{align} \]

\[\begin{align} f_{x} & = 2x + 2 = \lambda g_{x} = \lambda 2x \\ f_{y} & = 2y + 2 = \lambda g_{y} = \lambda 2y \\ \end{align} \]

This implies by rearranging terms:

\[ \begin{align} 1 = (\lambda -1 )x \\ 1 = (\lambda - 1 )y \\ x = y \text{ provided } \lambda \neq 1 \\ \end{align} \] That implies the critical points exist only if \(x = y\) and \(g(x,y) = 0\). \[g(x,y) = x^2 + y^2 - 4 = 0 \implies x = y = \sqrt{2}, -\sqrt{2}\].

Thus, the critical points exist at \(A = (\sqrt{2}, \sqrt{2})\) or \(B = (-\sqrt{2}, -\sqrt{2})\). Evaluating \(f\) at each of the two critical points gives:

\[ \begin{align} f(A) & = f(\sqrt{2}, \sqrt{2}) \\ & = (\sqrt{2})^2 + (\sqrt{2})^2 +2(\sqrt{2}) + 2(\sqrt{2}) \\ & = 4 ( 1 + \sqrt{2} ) \\ & \approx 9.656 \\ \end{align} \] \[ \begin{align} f(B) & = f(-\sqrt{2}, -\sqrt{2}) \\ & = (-\sqrt{2})^2 + (-\sqrt{2})^2 +2(-\sqrt{2}) + 2(-\sqrt{2}) \\ & = 4 ( 1 - \sqrt{2} ) \\ & \approx -1.656 \\ \end{align} \]

By comparing the values of the three critical points, we conclude the absolute minimum occurs at \(P = (-1,-1)\) with a value of \(-2\) and the absolute maximum occurs at \(A=(\sqrt{2}, \sqrt{2})\) with a value of \(9.656...\).

Graphical Solution

We verify the solution by graphing the function over a region containing the circle.

It is clear the graph is of a paraboloid whose central axis passes through (-1,-1). The absolute mimimum occurs at the bottom of the paraboloid over the central axis. The boundary of the function along the circle is an ellipse. The absolute maximum intuitively occurs on the boundary.