Sample Optimization Problem

Sample Optimization Problem - x t,y t = 4cos 4 t 4cos 2 t...

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MATH150 Sample Optimization Problem 11/18/05 Problem : Find the global maxima und minima of the function f ( x,y ) = x 4 + x 2 y 2 - 4 x on the disk x 2 + y 2 2 (This was the intended problem; I accidentally used x 2 + y 2 4 in class—this leads to a much harder problem) Solution : First we find the critical points of f : f ( x,y ) = (4 x 3 +2 xy 2 - 4 , 2 x 2 y ) = (0 , 0). This leads to the system I : 4 x 3 + 2 xy 2 - 4 = 0 II : 2 x 2 y = 0 Equation II yields x = 0 or y = 0. 1. Case: x = 0 If we plug this into equation I we get -4=0 which is a contradiction. We obtain no solution in this case. 2. Case: y = 0 If we plug this into equation I we get 4 x 3 - 4 = 0 x = 1. There is only one critical point: (1,0). There are no singular points. Now we consider the boundary: The boundary is a circle of radius 2. It can be parametrized as ( x ( t ) ,y ( t )) = ( 2cos t, 2sin t ) for 0 t < 2 π . We obtain g ( t ) = f (
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Unformatted text preview: x ( t ) ,y ( t )) = 4cos 4 t + 4cos 2 t sin 2 t-4 √ 2cos t = 4cos 2 t (cos 2 t + sin 2 t )-4 √ 2cos t = 4cos 2 t-4 √ 2cos t g ( t ) =-8sin t cos t + 4 √ 2sin t = 4sin t (-2cos t + √ 2) = 0 ⇔ sin t = 0 or cos t = √ 2 2 ⇔ t = 0 , t = π, t = π 4 , t = 7 π 4 We have 1 critical point and 4 boundary points to consider: g (0) = 4-4 √ 2, g ( π ) = 4 + 4 √ 2, g ( π/ 4) = g (7 π/ 4) = 4 · 1 2-4 √ 2 · √ 2 2 =-2, f (1 , 0) =-3. t = π corresponds to the point ( x ( π ) ,y ( π )) = (-√ 2 , 0). Thus, the absolute or global maximum of f is at (-√ 2 , 0) with a value of 4 + 4 √ 2; the absolute or global minimum of f is at (1 , 0) with a value of -3....
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