Find the absolute maximum and minimum values of f on the set D.f(x, y) = 5 + xy - x - 2y, &ensp; D is the closed triangular region with vertices (1, 0), (5, 0), and (1, 4). Created by Rain. science-mathematics-en - mathematics-en

Find the absolute maximum and minimum values of f on the set D. f(x, y) = 5 + xy - x

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Find the absolute maximum and minimum values of f on the set D. f(x, y) = 5 + xy - x - 2y, D is the closed t?

Find the absolute maximum and minimum values of f on the set D.

f(x, y) = 5 + xy - x - 2y, D is the closed triangular region with vertices (1, 0), (5, 0), and (1, 4)

D is bounded by the lines x = 1, y = 0, and y = -x + 5.

First of all, we find the critical points inside D.

Taking partial derivatives, f_x = y - 1, and f_y = x - 2.

Setting these equal to 0 yields (x, y) = (2, 1), which inside D (sketch it!).

Note that f(2, 1) = 3.

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Now, we check the boundary lines for extreme values.

(i) x = 1 ==> f(1, y) = 4 - y for y in [0, 4], which has extrema 4 (at y = 0) and 0 (at y = 4).

(ii) y = 0 ==> f(x, 0) = 5 - x for x in [1, 5], which has extrema 4 (at x = 1) and 0 (at x = 5).

(iii) y = -x + 5 ==> f(x, 5 - x) = -5 + 6x - x^2 = 4 - (x - 3)^2 for x in [1, 5],

which has extrema 4 (at x = 3) and 0 at (x = 1 or 5).

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Hence, the extreme values for f on D are 0 and 4.

I hope this helps!