Exam 1 - S06(1) - Math 10260 Exam 1 Solutions Spring 2006 1 Let us set 1980 to stand for t = 0 The total amount of oil consumed during 1980 to 2000 =

# Exam 1 - S06(1) - Math 10260 Exam 1 Solutions Spring 2006 1...

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Math 10260 Exam 1 - Solutions - Spring 2006 1. Let us set 1980 to stand for t = 0 . The total amount of oil consumed during 1980 to 2000 = The area under the graph [ r (5) + r (15)] Δ t = (6 + 8) · 10 = 140 . 2. Using integration by substitution u = ln x . Then du = 1 x dx , 2 1 ln x x dx = ln 2 0 udu . 3. Total change of profit = 30 20 [ - 0 . 5 q + 100] dq = - 0 . 5 2 [ q 2 + 100 q ] 30 20 = - 0 . 5 2 (30) 2 + 100(30) - - 0 . 5 2 (20) 2 + 100(20) = 875 . 4. CS = [ the area under the demand curve from 0 to q e ] – q e · p e = the area of the triangle = 1 2 · 10 · 2 = 10 . PS = q e · p e – [ the area under the supply curve from 0 to q e ] = 3 · 10 - 18 = 12. 5. Since g ( x ) f ( x ) on the interval [ - 2 , 1], then the area of the region between the curves of f ( x ) and g ( x ) = 1 - 2 [ g ( x ) - f ( x )] dx = 1 - 2 [2 - x - x 2 ] dx 6. Selling price = present value (PV) = 8 0 84000 e - 0 . 06 t dt = 84000 - 0 . 06 e - 0 . 06 t | 8 0 = 84000 - 0 . 06 [ e - 0 . 06 · 8 - 1] = \$533,703. 7. The rate of change of M = dM dt = rate due to interest – rate due to payment = 0 . 06 M - 9 , 600 . Being able to pay off the loan in 20 years means that; M (20) = 0 . 8. dp dt = 0 . 02 p - 0 . 004 p 2 = 0 . 02 p (1 - 0 . 004 0 . 02 p ) = 0 . 02 p (1 - 0 . 2 p ) = 0 . 02 p (1 - 1 5 p ) . Therefore K = 5 millions.
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