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C Roettger, Fall 14 Math 265 Quiz 7B – Solutions Problem 1 Consider the function f (x, y) = x2 y + 8x − y 3 + 2. Find all critical points of f (x, y) in the plane. Solution. First, compute the gradient of f (as always): ∇f (x, y) = (2xy + 8, x2 − 3y 2 ) This is the zero vector exactly if 2xy + 8 = 0 x2 − 3y 2 = 0. AND √ From the second equation, we get x = ± 3y. Substitute this into the first to get √ ±2 3y 2 + 8 = 0. Only the negative sign will give a solution, and for that case we can solve for y: s 2 4 = 2 · 3−1/4 . y = ± √ = ±√ 4 3 3 So we get two critical points (2 · 31/4 , 2 · 3−1/4 ) and (−2 · 31/4 , −2 · 3−1/4 ).