{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# CH12 - 1 MP CHAPTER 12 SOLUTIONS MP SECTION 12.1 1 lim h 0...

This preview shows pages 1–3. Sign up to view the full content.

MP CHAPTER 12 SOLUTIONS MP SECTION 12.1 3h + h 2 1. lim ---------- = lim (3 + h) = 3 h 0 h h 0 2. 25x 0 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 0 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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 41

CH12 - 1 MP CHAPTER 12 SOLUTIONS MP SECTION 12.1 1 lim h 0...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online