Mathematics 241, section 1 – Multivariable Calculus Solutions for Exam III – November 26, 2013

I. Given that

f(x, y) = \frac{y}{x^{2} + y^{2} + 4}
f_{x}(x, y) = \frac{−2xy}{x^{2} + y^{2} + 4)^{2}}
f_{y}(x, y) = \frac{x^{2} − y^{2} + 4}{(x^{2} + y^{2} + 4)^{2}}
f_{xx}(x, y) = \frac{2y(3x^{2} − y^{2} − 4)}{(x^{2} + y^{2} + 4)^{3}}
f_{xy}(x, y) = \frac{−2x(x^{2} − 3y^{2} + 4)}{(x^{2} + y^{2} + 4)^{3}}
f_{yy}(x, y) = \frac{−2y(3x^{2} − y^{2} + 12)}{(x^{2} + y^{2} + 4)^{3}},

find all critical points of f and apply the Second Derivative Test to each to determine the types of the critical points.

Solution: The critical points are the solutions of f_{x} = f_{y} = 0. From f_{x} = 0, we get either x = 0 or y = 0. There are no solutions of the equation f_{y} = 0 with y = 0, but if x = 0, we get y = ±2. So the critical points are (0, 2) and (0, −2). At (0, 2), f_{xx}(0, 2)f_{yy}(0, 2) − (f_{xy}(0, 2))^{2} = (−1/16)(−1/16) − 0^{2} = 1/256 > 0 and f_{xx}(0, 2) = −1/16 < 0. Therefore, f has a local maximum at (0, 2). At (0, −2), f_{xx}(0, −2)f_{yy}(0, −2) − (f_{xy}(0, −2))^{2} = (1/16)(1/16) − 0^{2} = 1/256 > 0. and f_{xx}(0, −2) = 1/16 > 0. Therefore f has a local minimum at (0, −2).

II.

A) (20) Find the maximum and minimum values of

f(x, y) = xy + 2y

on the closed and bounded set

S = {(x, y) ∈ ℝ^{2} | x^{2} + y^{2} ≤ 9}.

Solution: We have

f_{x} = y
f_{y} = x + 2