Calc III - 3.4-4.2 - Solutions.pdf - HW 7 3.4.4 Find the extrema of f(x y = x y subject to constraint x2 y 2 = 2 Solution rf =(1 1 rg =(2x 2y Local

Calc III - 3.4-4.2 - Solutions.pdf - HW 7 3.4.4 Find the...

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HW 7 3.4.4. Find the extrema of f ( x, y ) = x - y , subject to constraint x 2 - y 2 = 2. Solution: r f = (1 , - 1), r g = (2 x, - 2 y ) Local extrema must satisfiy r f = λ r g Solve it we get x = y, λ = 1 2 x put x=y into the constraint equation we have x 2 - x 2 = 0, which has no solution. So we conclude that there’re no extrema achieved under this constraint. 3.4.24. Find the absolute maximum and minimum value for f ( x, y, z ) = x + yz on the ball B = { ( x, y, z ) | x 2 + y 2 + z 2 1 } . Solution: r f = (1 , z, y ), which means r f 6 = 0 in the interior of the ball, so the candidates for absolute maximum and minimum can only happen on the boundary, which is @ B = { ( x, y, z ) | x 2 + y 2 + z 2 = 1 } under this constraint, r g = (2 x, 2 y, 2 z ), when achieving extrama, we have r f = λ r g . Solve it we get y = ± z, x = ± 1 or, so put it back to constraint equation @ B we get 1 + y 2 + z 2 = 1, so there’re 2 solutions: (1 , 0 , 0) and ( - 1 , 0 , 0). And these are the only points that can achive maximum or minimum. f (1 , 0 , 0) = 1 , f ( - 1 , 0 , 0) = - 1. So f achieve maximum at (1 , 0 , 0) with maximum value 1, achieves minimum at ( - 1 , 0 , 0) with minimum value - 1.
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