# h3s - 18.02 HOMEWORK#3 DUE OCTOBER 7 2010 BJORN POONEN 1...

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Unformatted text preview: 18.02 HOMEWORK #3, DUE OCTOBER 7, 2010 BJORN POONEN 1. Part A (After Sept. 24) 2F-1b, 2F-3* (12 pts.), 2F-4* (6 pts.) Solution. (to 2F-3) 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 x-intercept is the x-value when y = 0 and z = 0, which is (2 a + b- 1) /a . Similarly, the y-intercept is (2 a + b- 1) /b , and the z-intercept is 1- 2 a- b . The constraint inequalities on a and b are that these three intercepts are positive. But the z-intercept is- a times the x-intercept, and is- b times the y-intercept. 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 2F-4) 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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## This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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h3s - 18.02 HOMEWORK#3 DUE OCTOBER 7 2010 BJORN POONEN 1...

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