Math16B - Fall04 Final Solutions

# Math16B - Fall04 Final Solutions - SOLUTIONS TO MATH 16B...

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SOLUTIONS TO MATH 16B FINAL EXAM OF AUTUMN 2004 1. ∂f ∂x = 2 x - 2 y ∂f ∂y = - 2 x + 6 y + 4 Setting these equal to zero, we ﬁnd the only solution at ( - 1 , - 1). 2 f ∂x 2 = 2 2 f ∂y 2 = 6 2 f ∂x∂y = - 2 D f = 8 As D f > 0, there is a relative extremum at ( - 1 , - 1). As 2 f ∂x 2 > 2, this extremum is a minimum. Hence, the minimal value of f is f ( - 1 , - 1) = - 1 2. Z Z R ( y + ln( x )) dxdy = Z x = e x =1 Z 1 x y =0 ( y + ln( x )) dydx = Z x = e x =1 ( 1 2 y 2 + y ln( x )) | y = 1 x y =0 dx = Z x = e x =1 ( 1 2 x 2 + ln( x ) x ) dx = ( - 1 2 x + 1 2 (ln( x )) 2 ) | x = e x =1 = - 1 2 e + 1 2 + 1 2 = 1 - 1 2 e 3. Substitute u = e x + 2 so that du = e x dx . 1

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2 SOLUTIONS TO MATH 16B FINAL EXAM OF AUTUMN 2004 Z 0 e x e 2 x + 4 e x + 4 dx = lim r →∞ Z r 0 e x ( e x + 2) 2 dx = lim r →∞ Z u = e r +2 u =3 du u 2 = lim r →∞ - 1 u | u = e r +2 u =3 = lim r →∞ - 1 e r + 2 + 1 3 = 1 3 4. This problem was needlessly complicated. 5.
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Math16B - Fall04 Final Solutions - SOLUTIONS TO MATH 16B...

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