Final Exam - S09-1

# Final Exam - S09-1 - Math 10260 Final Exam Solutions Spring...

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Math 10260 Final Exam Solutions - Spring 2009 1. Since f ( x ) is a probability density function we have R 3 0 f ( x ) dx = 1 - 0 . 2 = 0 . 8. 2. P (3 < X < 4) = 0 . 2 3. The producer surplus (PS) is equal to the area of the rectangle [0 , 8] × [0 , 5] minus the area under the supply curve. That is, PS = 40 - R 8 0 S ( q ) dq . Using the indicated Riemann sum to estimate the integral in this formula we obtain PS = 40 - [2 · 2+(2 . 5) · 2+3 · 2+(3 . 75) · 2] = 40 - 22 . 50 = 17 . 5. 4. Equilibrium happens when demand equals suppl. Thus setting D ( q ) = S ( q ) gives 81 q +3 = q + 3, or ( q + 3) 2 = 81, or q + 3 = ± 9, or q = ± 9 - 3. Choosing the positive solution we obtain that q e = 6. Then, the equilibrium price is p e = S (6) = 6 + 3 = 9. 5. At any time t (during the duration of the mortgage) the rate of change of M ( t ) is equal to rate at which it increases due to interest minus the rate of payments. In symbols, this reads as: dM dt = 0 . 05 M - 18 , 000 . It is given that M (0) = 200 , 000. 6. We must compare the PV of option (I) with the amount offered in option (II). We have PV = Z 10 0 10 , 000 e 0 . 1 t e - 0 . 05 t dt = 10 , 000 Z 10 0 e 0 . 05 t dt = 10 , 000 e 0 . 05 t 0 . 05 10 0 = 200 , 000( e 0 . 5 - 1) 129 , 744 . Therefore option (I) is more beneficial. (Note: We could reach the same conclusion by computing FV’s.) 7. We have MPK = ∂P ∂K = 18 · 1 3 K - 2 3 L 2 3 , which for K = 1 , 000 and L = 27 , 000 is equal to 6(1 , 000) - 2 3 · (27 , 000) 2 3 = 3 · 900 100 = 54. 8. Let P ( x, y ) denote the profit at the production level ( x, y ). Then P ( x, y ) P (150 , 200) + ∂P ∂x (150 , 200)( x - 150) + ∂P ∂y (150 , 200)( y - 200) or P ( x, y ) 80 , 500+100( x - 150)+50( y - 200). Therefore P (160 , 190) 80 , 500+100(160 - 150) + 50(190 - 200) = 80 , 500 + 1 , 000 - 500 = 81 , 000.

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