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2014/15 AHSMC Part 1 Problem 15

Problem 15 of 16

For any real numbers xx and yy, the minimum value of x4−5x2+y2+2x+2y+2xy+6x^4 - 5x^2 + y^2 + 2x + 2y + 2xy + 6 is

Answer choices

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Solution

Since x4−5x2+y2+2x+2y+2xy+6=(x2−3)2+(x+y+1)2−4x^4 - 5x^2 + y^2 + 2x + 2y + 2xy + 6 = (x^2 - 3)^2 + (x + y + 1)^2 - 4, the minimum value is −4-4, attained at x=±3x = \pm\sqrt{3} and y=−1∓3y = -1 \mp \sqrt{3}. The answer is (B).

University of Alberta