Excercise_Solution_CH1

# Excercise_Solution_CH1 - order condition fails but there is...

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Solutions to Excercise in CH1 Slides Bin Xie 2/1/2015 Questions on Page 12. (1) y = ( x - 1)(5 - x ) Solution: dy dx = 1 · (5 - x ) + ( - 1) · ( x - 1) = 5 - x - x + 1 = 6 - 2 x use product rule for derivatives (2) y = x x - 3 + 5 x 3 Solution: dy dx = 1 · ( x - 3) - 1 · x ( x - 3) 2 + 15 x 2 = - 3 ( x - 3) 2 + 15 x 2 use quotient rule for derivatives. (3) y = x 3 (2 - 3 x ) 2 + lnx Solution: dy dx = 3 x 2 (2 - 3 x ) 2 + 2(2 - 3 x )( - 3) x 3 (2 - 3 x ) 4 + 1 x Questions on Page 20. (1) y = x 5 + lnx 5 - x Solution: dy dx = 5 x 4 + 1 x (5 - x ) - ( - 1) lnx (5 - x ) 2 = 5 x 4 + 5 x - 1 + lnx (5 - x ) 2 . (2) y = ( x - 1)( x - 2)( x - 3) Solution: y = ( x 2 - 3 x + 2)( x - 3) dy dx = (2 x - 3)( x - 3) + ( x 2 - 3 x + 2) = (2 x 2 - 9 x + 9) + ( x 2 - 3 x + 2) = 3 x 2 - 12 x + 11 . multiple ways to solve this one. (3) y = e ax 1 + bx 2 1

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Solution: ∂y ∂x 1 = ae ax 1 + bx 2 ; ∂y ∂x 2 = be ax 1 + bx 2 . 2 y ∂x 2 1 = a 2 e ax 1 + bx 2 ; 2 y ∂x 1 ∂x 2 = abe ax 1 + bx 2 ; 2 y ∂x 2 2 = b 2 e ax 1 + bx 2 . 2 y ∂x 2 ∂x 1 = abe ax 1 + bx 2 . (4) y = alnx 1 + blnx 2 Solution: ∂y ∂x 1 = a x 1 ; ∂y ∂x 2 = a x 2 2 y ∂x 2 1 = - a x 2 1 ; 2 y ∂x 1 ∂x 2 = 0 ; 2 y ∂x 2 2 = - a x 2 2 ; 2 y ∂x 2 ∂x 1 = 0 (5) y = - ( x - 3) 4 + 6 Solution: F.O.C. dy dx = - 4( x - 3) 3 = 0 yields a critical value x * = 3 . S.O.C. d 2 y dx 2 = - 12( x - 3) 2 = 0 . No maximum value under second order condition. (Actually this is a special case where second
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Unformatted text preview: order condition fails but there is clearly a maximized value y * = 6 . Ignore this example and stick to our F.O.C and S.O.C) (6) y = lnx + lny subject to 5 x + 3 y = 30 . Solution: (1) L = lnx + lny + λ (30-5 x-3 y ) F.O.C.: ∂ L ∂x = 1 x-5 λ = 0 ; ∂ L ∂x = 1 y-3 λ = 0 ; ∂ L ∂λ = 30-5 x-3 y = 0 . Solve the two ﬁrst order conditions: x = 1 5 λ and y = 1 3 λ . Plug into the third one: 30-2 λ = 0 . λ = 1 15 . So x = 3 and y = 5 . (2) y = lnx + ln ( 30-5 x 3 ) F.O.C. dy dx = 1 x + 3 30-5 x (-5 3 ) = 0 . So x = 3 and y = 5 . 2...
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