Find the maximum and minimum values of f(x,y) = x^4 + y^4 - 4xy + 2, on the rect

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First, findinding the critical points inside (or on) D: f_x = 4x^3 - 4y, f_y = 4y^3 - 4x. Set these equal to 0: y = x^3, and x = y^3 ==> x = (x^3)^3 = x^9 ==> x(x^8 - 1) = 0 ==> x = 0, -1, 1 (all other roots are not real). Since x must be in [0, 3], ignore x = -1. ==> (x, y) = (0, 0), (1, 1) are the critical points inside or on D. Note that f(0, 0) = 2, and f(1, 1) = 0. ------------------------- Now, we check the boundary of D. (i) x = 0 with y in [0, 2]: ==> f(0, y) = y^4 + 2, which has its extrema at the endpoints. ==> f(0, 0) = 2, and f(0, 2) = 18. (ii) x = 2 with y in [0, 2]: ==> f(2, y) = y^4 - 8y + 18 Setting f ' = 4y^3 - 8 = 0 ==> y = 2^(1/3) Check endpoints and this critical point: f(2, 0) = 18, f(2, 2) = 2, f(2, 2^(1/3)) = 18 - 6 * 2^(1/3) (iii) y = 0 with x in [0, 3]: ==> f(x, 0) = x^4 + 2, which has its extrema at the endpoints. ==> f(0, 0) = 2, f(3, 0) = 83. (iv) y = 2 with x in [0, 3]: ==> f(x, 2) = x^4 - 8x + 18, which has its extrema at the endpoints. Setting f ' = 4x^3 - 8 = 0 ==> x = 2^(1/3). Check endpoints and this critical point: f(0, 2) = 18, f(3, 2) = 75, f(2^(1/3), 2) = 18 - 6 * 2^(1/3). ------------------- From all of the points above, we see that we have a maximum of f(3,0) = 83 and a minimum of f(1,1) = 0.

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