Find the absolute maximum and minimum values of f on the set D. f(x, y) = x^2 +

Question

Find the absolute maximum and minimum values of f on the set D. f(x, y) = x^2 + y^2 - 2x, D is the closed triangular region with vertices (2, 0), (0, 2), and (0, -2) Since f is a polynomial it is continuous on D, so an absolute maximum and minimum exist. Here f_x = 2x - 2, f_y = 2y, and setting f_x = f_y = 0 gives (1, 0) as the only critical point (which is inside D), where f(1, 0) = -1. Along L_1: x = 0 and f(0, y) = y^2 for -2 lessthanorequalto y lessthanorequalto 2, a quadratic function which attains its minimum at y = 0, where f(0, 0) = 0. and its maximum at y = plusminus 2, where f(0, plusminus 2) = 4. Along L_2: y = x - 2 for 0 lessthanorequalto x lessthanorequalto 2, and f(x, x - 2) = 2x^2 - 6x + 4 = 2(x - 3/2)^2 -1/2, a quadratic which attains its minimum at x = 3/2, where f(3/2, - 1/2) = - 1/2, and it maximum at x = 0 where f(0, -2) = 4. Along L_a: y = 2 - x for 0 lessthanorequalto x lessthanorequalto 2, and f(x, 2 - x) = 2x^2 - 6x + 4 = 2(x - 3/2)^2 - 1/2, a quadratic which attains its minimum at x = 3/2, where f (3/2, 1/2) = - 1/2, and its maximum at x = 0, where f(0, 2) = 4. Thus the absolute maximum of f on D is f(0, plusminus 2) = 4 and the absolute minimum is f(1, 0) = -1.

Explanation / Answer

The function f has y^2 so the negative values of y does not affecting value of f. Therefore the maximum value of f depends on the magnitude of y but not its sign.

This the maximum value takes place at (0,-2).

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