^(-y-'h) (-!)-'(-5)- i- i

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1 68 Chapter 1 LINEAR FUNCTIONS The slope indicates that tuition and fees have increased approximately $1033 per year. (c) The year 202 is too far in the future to rely on this equation to predict costs; too many other factors may influence these costs by then. 1.2 Linear Functions and Applications 1. /(2) = 7 - (2) = 7-10 = /(4) = 7- (4) = 7-20 = /(-3) = 7 - (-3) = = /(-I) = 7 - (-l) = 7 + = 12. ^(1.) = 2(1.)-3 = 3-3 = 0 6. g(2.) = (2.) -3 = -3 = 2 ^(-y-'h) (-!)-'(-)- i- i 9. f(t) = 7- (0 = 7-* 10. g(k2) = 2{k2) - 3 = 2A: This statement is true. When we solve y = f(x) = 0, we are finding the value of x when y = 0, which is the x-intercept. When we evaluate /(0), we are finding the value of y when x = 0, which is the y-intercept. 12. This statement is false. The graph of f(x) = is a horizontal line. 13. This statement is true. Only a vertical line has an undefined slope, but a vertical line is not the graph of a function. There fore, the slope of a linear function cannot be de fined. 14. This statement is true. For any value of a, /(0) = a 0 = 0, so the point (0,0), which is the origin, lies on the line. 1. The fixed cost is constant for a particular prod uct and does not change as more items are made. The marginal cost is the rate of change of cost at a specific level of production and is equal to the slope of the cost function at that specific value; it approximates the cost of producing one additional item. 19. $10is the fixed cost and $2.2 is the cost per hour. Let x = number of hours; R(x) = cost of renting a snowboard for Thus, x hours. R(x) = fixed cost + (cost per hour) (number of hours) R(x) = 10 + (2.2)(ar) = 2.2x $10 is the fixed cost and $0.99 is the cost per down loaded song the marginal cost. Let x = the number of downloaded songs and C(x) = cost of downloading x songs Then, C{x) = (marginal cost) (number of downloaded songs) -f fixed cost C(x) = 0.99.T ^ is the fixed cost and 3^ is the cost per halfhour. Let x = the number of half-hours; C(x) = the cost of parking a car for Thus, x half-hours. C(x) = T = 3.T -I $44 is the fixed cost and $0.28 is the cost per mile. Let x = the number of miles; R(x) = the cost of renting for x Thus, miles. R(x) = fixed cost 4- (cost per mile) (number of miles) R(x) = r.

2 Section 1.2 Linear Functions and Applications Fixed cost, $100; 0 items cost $1600 to produce. Let C{x) = cost of producing x items. C(x) = mx + b, where b is the fixed Now, cost. C(x) = mx C{x) = 1600 when x = 0, so 1600 = m(0) = 0m 30 = m. Thus, C(x) = 30x Fixed cost: $3; 8 items cost $39. Let C{x) = cost of x items C(x) = mx + 6, where b is the fixed cost C(x) = mx D{q) = g (a) D(0) = (0) = 16-0 = 16 When 0 watches are demanded, the price is $16. (b) D{A) = (4) = 16- = 11 When 400 watches are demanded, the price is $11. (c) D(8) = (8) = = 6 When 800 watches are demanded, the price is $6. (d) Let D(q) = 8. Find q. 8 = ? -, = 8 q = 6A When the price is $8, 640 watches are demanded. (e) Let D(q) = 10. Find q. 10 = ? Now, C(x) = 39 when x = 8, so 39 = m(8) = 8m 4 = m. Thus, C(x) = 4.x Marginal cost: $7; 0 items cost $4300. C{x) = 7x + b Now, C{x) = 4300 when x = = 7(0) + b 4300 = b 0 = 6? = 4.8 When the price is $10, 480 watches are demanded. (f) Let D(q) = 12. Find q. 12 = ?? = 3.2 When the price is $12, 320 watches are demanded. (g) 16-k p ? Thus, C{x) = 7rr Marginal cost, $120; 700 items cost $96,00 to pro duce. C{x) = 120.x- + b Now, C{x) = 96,00 when x = ,00 = 120(700) + b 96,00 = 84,000 + b 12,00 = 6 Thus, C{x) = 120aH-12,00. (h) S(q) = 0.7? 1 I I I I M n q Let S(q) = 0. Find?. 0 = 0.7? 0 =? When the price is $0, 0 watches are supplied.

3 70 Chapter 1 LINEAR FUNCTIONS (i) Let S{q) = 10. Find?. 10 = 0.7? 40 T = q q = 13.3 When the price is $10, about 1333 watches are supplied. (j) Let S(q) = 20. Find?. 20 = 0.7? 80 T = q q = 26.6 When the price is $20, about 2667 watches are demanded. oo P=l6-\.2q (d) Let D(q) = 4.. Find?. 4. = - 0.2? 0.2? = 0.? = 2 When the price is $4.0, 200 quarts are demanded. (e) Let D{q) = 3.2. Find?. 3.2 = - 0.2? 0.2? = 1.7? = 7 When the price is $3.2, 700 quarts are demanded. (f) Let >(?) = 2.4. Find?. 2.4 = - 0.2? 0.2? = 2.6? = 10.4 When the price is $2.40, 1040 quarts are demanded «, 6- -.p**-0.2q q (1) D(q) = S(q) ? = 0.7? 16 = 2? 8 =? (8) = 0.7(8) = 6 The equilibrium quantity is 800 watches, and the equilibrium price is $ D(q) = - 0.2? (a) >(0) = - 0.2(0) =-0 = When 0 quarts are demanded, the price is $. (b) >(4) = - 0.2(4) = -1 = 4 When 400 quarts are demanded, the price is $4. (c) jd(8.4) = - 0.2(8.4) = = 2.9 When 840 quarts are demanded, the price is $ n (h) S{q) = 0.2? Let S(q) = 0. Find?. 0 = 0.2?? = 0 When the price is $0, 0 quarts are supplied, (i) Let S(q) = 2. Find?. 2 = 0.2?? = 8 When the price is $2, 800 quarts are supplied, (j) Let S(q) = 4.. Find?. 4. = 0.2??= 18 When the price is $4.0, 1800 quarts are supplied.

4 Section 1.2 Linear Functions and Applications 71 (k) 30. (a) (1) D(q) = S(q) - 0.2? = 0.2? = 0.? 10 =? (10) = 0.2(10) = 2. The equilibrium quantity is 1000 quarts and the equilibrium price is $ p=s(q) =??; p=d(q) =100 -? (a) (b) (?)=p=1.4?-0.6 D(q) =p=-2q Set supply equal to demand and solve for?. 1.4? = -2? ?+ 2? = ? = q=3a g«1.12 (1.12) = 1.4(1.12) -.6 = The equilibrium quantity is about 1120 pounds; the equilibrium price is about $ (a) C(x) = mx + b; m = 3.0; C(60) = 300 C{x) = 3.0a:+ b Find b. 300 = 3.0(60) + b 300 = = 6 C(x) = 3.0.x+ 90 (b) R(x) = 9.x C{x) = R.(x) (b) S(q) = D(q) 2 2,=100? = 12 (12) =?(12) =0 3.0a; + 90 = 9.x 90 =..x =.x Joanne must produce and sell 17 shirts. (c) P(.x) = R(x) - C(x); P(x) = = 9.x - (3.0x + 90) 00 =.a: =.x = x The equilibrium quantity is 12, the equilibrium price is $0 To make a profit of $00, Joanne must produce and sell 108 shirts.

5 72 Chapter 1 LINEAR FUNCTIONS 32. (a) C(x) =mx + b C(1000) = 267; b = 2 Find m. (b) R(x) = 4.9a: C(x) = R(x) 267 = m(1000) = 1000m 2.1 = in C(.x) = 2.1a: x+ 2 = 4.9.x 2 = 2.80a; 187. = x In order to break even, he must produce and sell 188 books. (c) P(x) = R(x) - C(x); P(x) = = 4.9.x - (2.1.x + 2) 1000 = 4.9.x-2.1.x = 2.80.x = 2.80.x 44.6 = x In order to make a profit of $1000, he must pro duce and sell 4 books. 33. (a) Using the points (100,11.02) and (400,40.12), m = y = 0.097(.x- 100) y = 0.097a; -9.7 y = x C{x) = x (b) The fixed cost is given by the constant in C(x). It is $1.32. (c) C(1000) = 0.097(1000) = = The total cost of producing 1000 cups is $ (d) C(1001) = 0.097(1001) = = = The total cost of producing 1001 cups is $98,417. (e) Marginal cost = = $0,097 or 9.7^ (f) The marginal cost for any cup is the slope, $0,097 or 9.7^. This means the cost of producing one additional cup of coffee would be 9.7yf. 34. C{10,000) =47,00; C(0,000) =737,00 (a) C(x) = mx + b m = 737,00-47,00 0,000-10, ,000 40,000 = 4.7 y- 47,00 = 4.7(.x - 10,000) y - 47,00 = 4.7.x - 47,00 y = 4.7a; + 00,000 C(x) = 4.7.x + 00,000 (b) The fixed cost is $00,000. (c) (7(100,000) = 4.7(100,000) + 00,000 = 47, ,000 = 97,000 The total cost to produce 100,000 items is $97,000. (d) Since the slope of the cost function is 4.7, the marginal cost is $4.7. This means that the cost of producing one additional item at this production level is $ (a) (100,000)(0) =,000,000 Sales in 1996 would be 100,000 +,000,000 =,100,000. (b) The ordered pairs are (1, 100,000) and (6,,100,000).,100, ,000,000,000 ftn_ m = = = 1,000, y - 100,000 = l,000,000(a- - 1) y - 100,000 = 1,000,000.x - 1,000,000 y= 1,000,000.x-900,000 (.x) = 1,000,000.x - 900,000 (c) Let (.x) = 1,000,000,000. Find.x. 1,000,000,000 = 1,000,000.x - 900,000 1,000,900,000 = 1,000,000.x x = Sales would reach $1 billion in about = , or during the year Sales would have to grow much faster than lin early to reach $1 billion by 2003.

6 Section 1.2 Linear Functions and Applications 73 (d) Use ordered pairs (13, 36,000,000) and (14, 479,000,000). m=479'000' -3f'000'000 =123,000,000 S{x) - 36,000,000 = 123,000,000(.x - 13) (.x) - 36,000,000 = 123,000,000.x - 1,99, S(x) = 123,000,000.x - 1,243,000,000 (e) The year 200 corresponds to x = = 1. (1) = 123,000,000(1) - 1,243,000,000 (1) = 602,000,000 The estimated sales are $602,000,000, which is less than the actual sales. (f) Let (a:) = 1,000,000,000. Find x. 1,000,000,000 = 123,000,000a: - 1,243,000,000 2,243,000,000 = 123,000,000.x x «18.2 Sales would reach $1 billion in about = , or during the year C(x) =.x + 20; R(x) = 1a: (a) C(x) = R.(x) x + 20 = 1a; 20 = 10a; 2 = x The break-even quantity is 2 units. (b) P{x) = R{x) - C(x) P(x) = 1a;- (a;+ 20) P(100) = 1(100) - ( ) = = 980 The profit from 100 units is $980. (c) P(x) = 00 1x - (a; + 20) = 00 10a; - 20 = 00 10a; = 20.x = 2 For a profit of $00, 2 units must be produced. 37. C(x) = 12a- + 39; R(x) = 2.x (a) C(x) = R(x) 12.x + 39 = 2.x 39 = 13a: 3 =.x The break-even quantity is 3 units. (b) P(x) = R(x) - C(x) P(x) = 2.x - (12a: + 39) P(x) = 13a;- 39 P(20) = 13(20) - 39 = = 3211 The profit from 20 units is $3211. (c) P(x) = $130; find a;. 130 = 13.x = 13a; 13 =.x For a profit of $130, 13 units must be produced. 38. C(x) = 8a;+ 900 R{x) = 10.x Set C(x) = R(x) to find the break-even quantity. 8.x = 10.x 900 = 20a; 4 =.x The break-even quantity is 4 units. You should decide not to produce since no more than 38 units can be sold. P{x) = R(x) - C{x) = 10a: - (8a: + 900) = 20a: The profit function is P(x) = 20a: C{x) = 10.x R(x) = 20.x Set C(x) = R(x) to find the break-even quantity. 10a; = 20.x 6000 = 14.x w x The break-even quantity is about 41 units, so you should decide to produce. P(.x) = #(.x) - C(x) = 20a:-(10.x+ 6000) = 14a: The profit function is P(.x) = 14a:

7 74 Chapter 1 LINEAR FUNCTIONS 40. C(x) = 70.x + 00 R(x) = 60a; 70a; + 00 = 60.x 10a; = -00 x = -0 This represents a break-even quantity of 0 units. It is impossible to make a profit when the break even quantity is negative. Cost will always be greater than revenue. P(x) = R(x) - C(.x) = 60a: - (70.x + 00) =-10.x-00 The profit function is P(x) = -10a: C(x) = 1000.x R(x) = 900.x 900.x = 1000.x = 100a; -0 = x It is impossible to make a profit when the break even quantity is negative. Cost will always be greater than revenue. P(x) = R{x) - C(x) = 900a; - (1000.x + 000) =-100.x-000 The profit function is P(x) = 100a; 000 (always a loss). 42. Use the formulas derived in Example 7 in this sec tion of the textbook. (a) F = 8; find C. C= (8-32) C=(26) C = 14.4 P=?C +32 C=^(F-32) The temperature is 14.4 C. (b) F = -20; find C. C= (F-32) C=jj(-20-32) C= (-2) C = The tempereature is 28.9 C. (c) C = 0; find F. F=^C+'S2 ^ =1(0) +32 F = F=122 The temperature is 122 F. 43. Use the formula derived in Example 7 in this sec tion of the textbook. F=?C+32 C= (F-32) (a) C = 37; find F. F= j?(37) +32 F=^+32 F = 98.6 The Fahrenheit equivalent of 37 C is 98.6 F. (b) C = 36.; find F. F= (36.)+32 F = F = 97.7 C = 37.; find F. F=?(37.) +32 = = 99. The range is between 97.7 F and 99. F.

8 Section 1.3 The Least Squares Line If the temperatures are numerically equal, then F = C. F= c+32 C=?C ^C = 32 C=-40 The Celsius and Fahrenheit temperatures are nu merically equal at The Least Squares Line 2. For the set of points (1,4), (2,), and (3,6), Y = x + 3. For the set (4,1), (,2), and (6,3), Y = x - 3. (c) The least squares line is of the form Y = m.x + b. First solve for m. Now find b. m = b = n( *y)-(e*)(ey)»( **)-G»2 10(186) - ()(2.) 10(38) - () «0. E V~ m(e x) n () = -0. Thus, Y = 0..x (a) y I I I I I I I I I (b) * y xy x2 y (d) Let x = 11. Find Y. Y = 0.(11) -0. =. 4..X y xy a-2 y r = (37.81) - (3.)(.3) ^(22.2) - (3.)2 ^(.9) - (.3)2 r2 = (0.698)2 «0. «0.698 r = (E :cy)-(z»(ea/) \ME*2) - Q»2 MEy2) - (Es/)2 10(186) - ()(2.) ^10(38) - ()2^10(90.7) - (2.)2 The answer is choice (c)