Let fx, yx 2 y 2 2x 6y 14 a, ba . Then fxx, y2x 2 fyx, y2y
Chapter 14, Problem 14.427(choose chapter or problem)
Let fx, yx 2 y 2 2x 6y 14 a, ba . Then fxx, y2x 2 fyx, y2y 6 fx, yx These partial derivatives are equal to 0 when and , so the only critical point is . By completing the square, we find that fx, y4 x 1 2 y 3 2 1, 3 x 1 Since and , we have for all values of and . Therefore is a local minimum, and in fact it is the absolute minimum of . This can be confirmed geometrically from the graph of which is the elliptic paraboloid with vertex shown in Figure 2
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