h4s - 18.02 HOMEWORK #4, DUE OCTOBER 14, 2010 BJORN POONEN...

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Unformatted text preview: 18.02 HOMEWORK #4, DUE OCTOBER 14, 2010 BJORN POONEN In the Lagrange multiplier problems, don’t forget to check all the things you are supposed to check. 1. Part A (After Oct. 5) 2I-1b, 2I-3 (After Oct. 5) p .924: 36* (15 pts.) (Heron’s formula should read A = p s ( s- x )( s- y )( s- z )), 38* (15 pts.) Solution. (to p. 924, 36) Given the perimeter 2 s , we are trying to maximize the function f ( x, y, z ) := A 2 = s ( s- x )( s- y )( s- z ) subject to the constraint equation g ( x, y, z ) := x + y + z = 2 s , and the constraint inequalities x > 0, y > 0, z > 0, x < y + z , y < x + z , z < x + y (the necessary and sufficient conditions for there to exist a triangle with side lengths x, y, z ). We compute ∇ f = h- s ( s- y )( s- z ) ,- s ( s- x )( s- z ) ,- s ( s- x )( s- y ) i ∇ g = h 1 , 1 , 1 i . Next we solve the system x + y + z = 2 s h- s ( s- y )( s- z ) ,- s ( s- x )( s- z ) ,- s ( s- x )( s- y ) i = λ h 1 , 1 , 1 i . The latter implies that- s ( s- y )( s- z ) =- s ( s- x )( s- z ) =- s ( s- x )( s- y ) . Dividing by- s ( s- x )( s- y )( s- z ) (the constraint inequalities guarantee that this is nonzero) yields 1 s- x = 1 s- y = 1 s- z . Taking reciprocals yields s- x = s- y = s- z, which implies x = y = z . This, together with x + y + z = 2 s , implies x = y = z = 2 s/ 3, and corresponds to the equilateral triangle. To finish the problem, we check • Points on g = c (i.e., x + y + z = 2 s ) where f or g is not differentiable. (There are none.) • Points where ∇ g = 0. (There are none.) • Behavior as ( x, y, z ) tends to ∞ . (Not applicable, because the constraints define a bounded region.) 1 • Boundary points. As ( x, y, z ) approaches the boundary of the region defined by the constraint inequalities, the triangle is appraoching a degenerate triangle with one side equal to 0, or x + y = z (a “flat triangle” with the three vertices along a line), or the same with x, y, z permuted. In all these cases, the area is 0, but the equilateral triangle found above has area larger than 0. Thus the equilateral triangle has the maximum area, among all triangles with the given perimeter, 2 s . Solution. (to p. 924, 38) We want to maximize/minimize the function f ( x, y ) := x 2 + y 2 subject to the constraint equation g ( x, y ) := x 2 + xy + y 2 = 3. We compute ∇ f = h 2 x, 2 y i ∇ g = h 2 x + y, x + 2 y i , so the Lagrange multiplier equations are x 2 + xy + y 2 = 3 2 x = λ (2 x + y ) 2 y = λ ( x + 2 y ) ....
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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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h4s - 18.02 HOMEWORK #4, DUE OCTOBER 14, 2010 BJORN POONEN...

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