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Unformatted text preview: 18.02 HOMEWORK #3, DUE OCTOBER 7, 2010 BJORN POONEN 1. Part A (After Sept. 24) 2F1b, 2F3* (12 pts.), 2F4* (6 pts.) Solution. (to 2F3) The plane has the form z = ax + by + c . Since (2 , 1 , 1) is on it, 1 = 2 a + b + c , so c = 1 2 a b . Thus the plane is z = ax + by +(1 2 a b ). The xintercept is the xvalue when y = 0 and z = 0, which is (2 a + b 1) /a . Similarly, the yintercept is (2 a + b 1) /b , and the zintercept is 1 2 a b . The constraint inequalities on a and b are that these three intercepts are positive. But the zintercept is a times the xintercept, and is b times the yintercept. So to say that the three intercepts are positive is the same as requiring that a and b are negative, and 1 2 a b is positive. Actually the last condition is a consequence of the first two, so the only constraint inequalities are that a and b are negative; i.e., ( a, b ) is in the third quadrant. The function to be minimized is the product f ( a, b ) := 2 a + b 1 a 2 a + b 1 b (1 2 a b ) = (1 2 a b ) 3 ab . The numerator is at least max(  a  ,  b  ) 3 , but the denominator is at most max(  a  ,  b  ) 2 , so f ( a, b ) ≥ max(  a  ,  b  ), which goes to ∞ as either  a  or  b  grows. Also as ( a, b ) approaches either axis (within the third quadrant), f ( a, b ) tends to ∞ . Thus the minimum of f ( a, b ) will occur somewhere inside at the third quadrant. Since f is differentiable everywhere inside the third quadrant, the minimum must occur at a critical point of f . To find the critical points of f with a, b < 0, we calculate ∂f ∂a = ab (3(1 2 a b ) 2 ( 2)) (1 2 a b ) 3 b ( ab ) 2 = (1 2 a b ) 2 ( 6 a (1 2 a b )) a 2 b ∂f ∂b = ab (3(1 2 a b ) 2 ( 1)) (1 2 a b ) 3 a ( ab ) 2 = (1 2 a b ) 2 ( 3 b (1 2 a b )) ab 2 . The factor 1 2 a b is positive, so if both these derivatives are zero, then the other factors in the numerator must be zero: 6 a (1 2 a b ) = 0 3 b (1 2 a b ) = 0 . These imply 6 a = 3 b so b = 2 a . Substituting into the first equation yields 6 a (1 2 a 2 a ) = 0, which becomes 2 a 1 = 0, so a = 1 / 2 and b = 2 a = 1. Thus the only critical point is at ( a, b ) = ( 1 / 2 , 1), so it must be the minimum that we knew existed. The answer to the problem, the corresponding plane, is z = 1 2 x y + 3 . 1 Solution. (to 2F4) As a polynomial in x (with coefficients that are polynomials in y ), the function is x 2 + (2 y ) x + (4 y 2 + 6) . Completing the square rewrites this as ( x + y ) 2 y 2 + (4 y 2 + 6) = ( x + y ) 2 + 3 y 2 + 6 ....
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 Fall '08
 Auroux
 Critical Point, Multivariable Calculus, BJORN POONEN

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