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Problem%20set2

# Problem%20set2 - L = 10 and K = 5 then ln Q = 4 872 and Q =...

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1 Problem set 2 3.1.6 P 2001 = \$50 million ° (1 : 06) = \$53 million P 2003 = \$50 million ° (1 : 06) 3 = \$59 : 55 million P 1998 = \$50 million ° (1 : 06) ° 2 = \$44 : 50 million 3.2.8 At 7% PV = 15000 ° e ° 0 : 07 = 13 ; 986 At 5% PV = 15000 ° e ° 0 : 05 = 14 ; 268 At 9.5% PV = 15000 ° e ° 0 : 095 = 13 ; 641 3.3.4 Option 1 : Keep the collection which ensures ypu \$3000 after ten years. The cost of this option is the cost fee rising at the continuosly compounded rate of 5%. Cost ( at the end of 10th year) = 200 ° e 0 : 05 ± 10 = \$329 : 75 Net _ V alue ( at the end of 10th year) = 3000 ± 329 : 75 = \$2670 : 25 Option 2 : Invest in the stock market, which yields a value of V alue ( at the end of 10th year) = 1000 ° e 0 : 095 ± 10 = \$2585 : 71 Result: Based on this information keep the collection ! 3.3.10 1

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Given the Cobb-Douglas production function: Q = 15 L 4 = 5 K 1 = 5 Transform the above fundtion using natural logarithms into a linear function: ln Q = ln(15) + 4 5 ln( L ) + 1 5 ln( K ) If L = 10 and K = 5 , then
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Unformatted text preview: L = 10 and K = 5 , then ln Q = 4 : 872 and Q = exp(ln Q ) = 130 : 6 4.1.2 a) Set use the condition x = y to get: 6 z + 3 h & 4 a + 10 = 4 z & h + 6 & z = 2 a & 2 h & 2 Use the second equation to &gure out & y: & y = 4& z & h + 6 & y = 8 a & 9 h & 2 Then using the third equation we have & x = & y Hence, & z = 2 a & 2 h & 2 & y = 8 a & 9 h & 2 & x = 8 a & 9 h & 2 b) 2 & x = 8& a & y = 8& a & z = 2& a c) & x = 16 & y = 16 & z = 4 6.2.8 For x = 3 and & x = 3 ; then using the formula for the di/erence quotient found in the book on page 150, & y & x = 34 a) When & x = 1 : 5 then & y & x = 28 When & x = 0 : 5 then & y & x = 24 b) In the milit, as & x approaches zero, & y & x = 22 3...
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