Math16B - Spring 08 MT1 Solutions

Math16B - Spring 08 MT1 Solutions - 4. F ( x, y ) = 2 ye...

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SOLUTIONS TO SPRING 2008 MIDTERM 1 FOR MATH 16B 1. F ( s, t, u ) = s 2 e t 2 u + s 4 + ln( u 3 + st ) ( s 2 + e s ) ∂F ∂u = t 2 s 2 e t 2 u + 3 u 2 ( u 3 + st )( s 2 + e s ) hence 2 F ∂u∂t = ∂t ( ∂F ∂u ) = 2 ts 2 e t 2 u + 2 t 3 us 2 e t 2 u + - 3 su 2 ( u 3 + st ) 2 ( s 2 + e s ) 2. Using the method of Lagranage multipliers, the associated function is H ( x, y, z, λ ) = x 2 + 3 y 2 + z 2 - 2 x - 5 y - 4 z + 10 + λ (3 x + 2 y + 4 z - 15) Taking each partial derivative and setting the results equal to zero we have 0 = ∂H ∂x = 2 x - 2 + 3 λ 0 = ∂H ∂y = 6 y - 5 + 2 λ 0 = ∂H ∂z = 2 z - 4 + 4 λ 0 = ∂H ∂λ = 3 x + 2 y + 4 z - 15 Combining the first two equations, eliminating λ , we find x = 9 2 y - 11 4 The second two equations yield z = 6 y - 3 Substituting into the last equation, we have 79 2 y = 141 4 or y = 141 158 Thus, x = 100 79 and z = 186 79 1
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2 SOLUTIONS TO SPRING 2008 MIDTERM 1 FOR MATH 16B 3. The sum of squares of errors for f is ( 3 2 - 1) 2 + (3 - 2) 2 + ( 7 2 - 3) 3 = 1 . 5 while the sum of squares of errors for g is ( - 1 - 1) 2 + (2 - 2) 2 + (3 - 3) 3 = 4. Thus, f is a closer fit (based on the least squares method).
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Unformatted text preview: 4. F ( x, y ) = 2 ye x-32 x-y 2 + 5 hence F x = 2 ye x-32 F y = 2 e x-2 y Setting these equal to zero, using F y = 0 we nd y = e x . Substituting into the rst equation, we nd 2 e 2 x = 32. Dividing by 2 and taking natural logarithms, we nd x = ln(4) and thus, y = 4. Computing second derivatives, 2 F x 2 = 2 ye x 2 F y 2 =-2 2 F xy = 2 e x Hence, the Hessian is D F =-4 ye x-4 e 2 x So that at the one potential relative extremum or saddle point, we have D F (ln(4) , 4) =-128 < 0. Hence, this point is a saddle point. 5. a. A,D,F; b. H; c. B; d. K; e. I...
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This note was uploaded on 05/12/2010 for the course CHEM 3A taught by Professor Fretchet during the Fall '08 term at University of California, Berkeley.

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Math16B - Spring 08 MT1 Solutions - 4. F ( x, y ) = 2 ye...

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