CH12 - MP CHAPTER 12 SOLUTIONS MP SECTION 12.1 3h + h 2 1....

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Unformatted text preview: MP CHAPTER 12 SOLUTIONS MP SECTION 12.1 3h + h 2 1. lim ---------- = lim (3 + h) = 3 h → 0 h h → 2. 25x ≤ x ≤ 100 f(x) = 2500 + 20(x - 100) 100 ≤ x ≤ 200 25(100) + 20(100) + 15(x - 200) x ≥ 200 This function is continuous for all non-negative x. However, f(x) has no derivative at x = 100 and x = 200 (the slope of f(x) abruptly changes at these points). 3a. x(-e-x ) + e-x (x 2 + 1)(2x) -x 2 (2x) 3b. ---------------------- (x 2 + 1) 2 3c. 3e 3x 3d. -6/(3x + 2) 3 3e. 3x 2 /x 3 = 3/x 4. ∂ f/ ∂ x 1 = 2x 1 exp(x 2 ) ∂ f/ ∂ x 2 = x 1 2 exp(x 2 ) ∂ f 2 / ∂ x 1 ∂ x 2 = ∂ f 2 / ∂ x 2 ∂ x 1 = 2x 1 exp(x 2 ) = ∂ 2 f/ ∂ 2 x 1 = 2exp(x 2 ), ∂ 2 f/ ∂ 2 x 2 = x 1 2 exp(x 2 ) 5a. f'(p)<0 if a price increase lowers demand. Thus we expect to find f'(p)<0. 5b. Let r(p) = pf(p). If r'(p)<0, then price decrease increases revenue. Now r'(p) = pf'(p) +f(p)<0 if dq p --- < -q or dp p dq- --- < -1 or E<-1. q dp 5c. If -1<E<0, then r'(p)>0, so a price cut will decrease revenue. 6a. lim k( 1 - e-cx ) = k x →∞ 1 6b. The maximum size of the market as measured in terms of sales per year. 6c. If f(x) = k(1- e-cx ) is sales response from $x of advertising, then f'(x) is (approximately) the sales response due to increasing advertising from x to x + 1. Note that f'(x) = kce-cx = c times (part of market not buying product). 7a. Cost of producing x'th unit is c'(x) = k(1 - b)x-b which is a decreasing function of x. 7b. If total amount produced is x, then production cost per unit is k(1 - b)x-b . If total amount produced is 2x, then production cost per unit is k(1 - b)(2x)-b . Thus doubling the amount produced reduces cost per unit to 100(2-b )% of what it was previously.. 8. Let total output = f(m,w) = 3m 1/3 w 2/3 . Then ∂ f/ ∂ w = 2m 1/3 w-1/3 , ∂ f/ ∂ m = m-2/3 w 2/3 . Thus ∂ f/ ∂ m(216, 1000) = 1/36(100) = 100/36 ∂ f/ ∂ w(216, 1000) = 2(6)(1/10) = 1.2 One extra hour of machine time increases output by approximately 1(100/36) = 100/36 units while two hours of labor increases output by approximately 2(1.2) = 2.4 units. Thus one hour of machine time is a better buy than two hours of labor. MP SECTION 12.2 1a. Let S = soap opera ads and F = football ads. Then we wish to min z = 50S + 100F st 5S 1/2 + 17F 1/2 ≥ 40 (men) 20S 1/2 + 7F 1/2 ≥ 60 (women) S ≥ 0, F ≥ 1b. Since doubling S does not double the contribution of S to each constraint,we are violating the proportionality assumption. Additivity is not violated. 1c. This accounts for the fact that an extra soap opera ad yields a benefit which is a decreasing function of the number of football ads. This accounts for the fact that we may not want to double count people who see both types of ads....
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This note was uploaded on 10/31/2010 for the course PHYSICAL C 11209 taught by Professor Whaley during the Fall '10 term at University of California, Berkeley.

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CH12 - MP CHAPTER 12 SOLUTIONS MP SECTION 12.1 3h + h 2 1....

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