1.4 CONCEPT QUESTIONS, page 49

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1 .4 CONCEPT QUESTIONS, page 49. The intersection must lie in the first quadrant because only the parts of the demand and supply curves in the first quadrant are of interest.. a. The breakeven point P0( x0, p 0) is the solution of the simultaneous equations p = Rx ( ) and p = Cx ( ). b. The number x 0 is called the breakeven quantity. c. The number p 0 is called the breakeven revenue.. a. Market equilibrium occurs when the product produced is equal to the quantity demanded. b. The quantity produced at market equilibrium is called the equilibrium quantity. c. The price of the product produced at market equilibrium is called the equilibrium price. EXERCISES.4, page 49. We solve the system y = x + 4 y = x + 4. Substituting the first equation into the second yields x + 4 = x + 4 5x = 0, and x =. Substituting this value of x into the first equation yields y = () + 4, or y = 0. Thus, the point of intersection is (,0).. We solve the system y = 4x 7 y = 5x + 0. Substituting the first equation into the second yields ( 4x 7) = 5x + 0 4x + 7 = 5x + 0 x =. Substituting this value of x into the first equation, we obtain y = 4( ) 7 = 7 = 5. Therefore, the point of intersection is (,5).. We solve the system x y = 6 x + 6y = 6. Solving the first equation for y, we obtain 45 Straight Lines and Linear Functions

2 y = x 6 y = x. Substituting this value of y into the second equation, we obtain x+ 6( x ) = 6 x + 4x = 6 7x = 8 and x = 4. Then y = ( 4) =. Therefore, the point of intersection is ( 4, ). 4. We solve the system x + 4y = 5x + y = 5. Solving the first equation for x, we find x = y +. Substituting this value into the second equation of the system, we have 5( y+ ) + y = y + y = 5 0y y = 0 6y = 65, or y = 5/. Substituting this value of y into the first equation, we have x + 4( 5 ) = x = or x = /. Thus, the point of intersection is (, 5 ).. y = 4 x 5 5. We solve the system. Substituting the value of y given in the first x y = equation into the second equation, we obtain x ( 4 x 5) = 5 x 8 x+ = 6x x + 60 = 8 x = 5, or x = 4. Substituting this value of x in the first equation, we have y = 4 ( 4) 5 = 5, or y = 6. Therefore, the point of intersection is ( 4, 6). Straight Lines and Linear Functions 46

3 R S T 6. We solve the system y = x 4. Substituting the first equation into the second x+ y+ = 0 equation, we obtain x+ ( x 4) + = 0 x + x + = 0, x = 9, or x =. Substituting this value of x in the first equation, we have y = ( ) =. Therefore, the point of intersection is (, ). 7. We solve the equation R(x) = C(x), or 5x = 5x + 0,000, obtaining 0x = 0,000, or x = 000. Substituting this value of x into the equation R(x) = 5x, we find R(000) = 5,000. Therefore, the breakeven point is (000, 5,000). 8. We solve the equation R(x) = C(x), or x = 5x +,000, obtaining 6x =,000, or x = 000. Substituting this value of x into the equation R(x) = x, we find R(000) = 4,000. Therefore, the breakeven point is (000, 4,000). 9. We solve the equation R(x) = C(x), or 0.4x = 0.x + 0, obtaining 0.x = 0, or x = 600. Substituting this value of x into the equation R(x) = 0.4x, we find R(600) = 40. Therefore, the breakeven point is (600, 40). 0. We solve the equation R(x) = C(x) or 70x = 50x + 0,000, obtaining 500 0x = 0,000 or x = 67. Substituting this value of x into the equation R(x) = 70x, we find R(67) = 45,090. Therefore, the breakeven point is (67, 45,090).. a. b. We solve the equation R(x) = C(x) or 4x = 8x + 48,000, obtaining 6x = 48,000 or x = Substituting this value of x into the equation R(x) = 4x, we find R(8000) = 4(8000) =,000. Therefore, the breakeven point is (8000,,000). 47 Straight Lines and Linear Functions

4 c. d. P(x) = R(x) C(x) = 4x 8x 48,000 = 6x 48,000. The graph of the profit function crosses the x-axis when P(x) = 0, or 6x = 48,000 and x = This means that the revenue is equal to the cost when 8000 units are produced and consequently the company breaks even at this point.. a. R(x) = 8x; C(x) = 5,000 + x, P(x) = R(x) C(x) = 5x 5,000. Next, the breakeven point occurs when P(x) = 0, that is, 5x 5,000 = 0 or x = Then R(5000) = 40,000, so the breakeven point is (5000, 40,000). b. If the division realizes a 5 percent profit over the cost of making the diaries, then P(x) = 0.5 C(x), or 5x 5,000 = 0.5(5,000 + x); 4.55x = 8,750, and x = 68.68, or approximately 69 diaries.. Let x denote the number of units sold. Then, the revenue function R is given by R(x) = 9x. Since the variable cost is 40 percent of the selling price and the monthly fixed costs are $50,000, the cost function C is given by C(x) = 0.4(9x) + 50,000 =.6x + 50,000. To find the breakeven point, we set R(x) = C(x), obtaining 9x =.6x + 50, x = 50,000 x 959, or 959 units. Substituting this value of x into the equation R(x) = 9x gives R(959) = 9(959) = 8,. Thus, for a breakeven operation, the firm should manufacture 959 bicycle pumps resulting in a breakeven revenue of $8,. 4. a. The cost function associated with renting a truck from the Ace Truck Leasing Straight Lines and Linear Functions 48

5 Company is C (x) = x. The cost function associated with renting a truck from the Acme Truck Leasing Company is C (x) = x. b. c. The cost of renting a truck from the Ace Truck Leasing Company for one day and driving 70 miles is C (70) = (70) = 40.50, or $ The cost of renting a truck from the Acme Truck Leasing Company for one day and driving it 70 miles is C (70) = (70) = 9, or $9. Thus, the customer should rent the car from Acme Truck Leasing Company. This answer may also be obtained by inspecting the graph of the two functions and noting that the graph of C (x) lies below that of C (x) for x < a. The cost function associated with using machine I is given by C (x) = 8, x. The cost function associated with using machine II is given by C (x) = 5, x. b. c. Comparing the cost of producing 450 units on each machine, we find C (450) = 8, (450) 49 Straight Lines and Linear Functions

6 = 4,750 or $4,750 on machine I, and C (450) = 5, (450) = 4,000 or $4,000 on machine II. Therefore, machine II should be used in this case. Next, comparing the costs of producing 550 units on each machine, we find C (550) = 8, (550) = 6,50 or $6,50 on machine I, and C (550) = 5, (550) = 6,000 on machine II. Therefore, machine II should be used in this instance. Once again, we compare the cost of producing 650 units on each machine and find that C (650) = 8, (650) = 7,750, or $7,750 on machine I and C (650) = 5, (650) = 8,000, or $8,000 on machine II. Therefore, machine I should be used in this case. d. We use the equation P(x) = R(x) C(x) and find P(450) = 50(450) 4,000 = 500, or a loss of $500 when machine II is used to produce 450 units.similarly, P(550) = 50(550) 6,000 = 500, or a profit of $500 when machine II is used to produce 550 units. Finally, P(650) = 50(650) 7,750 = 4750, or a profit of $4750 when machine I is used to produce 650 units. 6. First, we find the point of intersection of the two straight lines. (This will give the time when the sales of both companies are the same). Substituting the first equation into the second gives t = t. = 0.t. and t = = 5 0. From the observation that the sales of Cambridge Drug Store is increasing at a faster rate than that of the Crimson Drug Store (its trend line has the greater slope), we conclude that the sales of the Cambridge Drug Store will surpass the annual sales of the Crimson Drug Store in 5½ years. 7. We solve the two equations simultaneously, obtaining Straight Lines and Linear Functions 50

7 8t+.4 = t+ 88 0t = 74.6 t.486 or approximately.5. So the shipments of LCDs will first overtake the shipments of CRTs a little before the middle of a. The number of digital cameras sold in 00 is given by f (0) =.05(0) = 6.85 or 6.85 million. The number of film cameras sold in 00 is given by g (0) =.85(0) Therefore, more film cameras than digital cameras were sold in 00. b. The sales are equal when.05t =.85t t = t = =.986 or approximately years. 4.9 Therefore, digital sales surpassed film sales near the end of a. b. We solve the two equations simultaneously. Thus, t + = t = 0 t = 4.09 The transactions done electronically will first exceed those done by check in early a. 5 Straight Lines and Linear Functions

8 b. 7.5t+ 5 =.9t+ 4.5,.4t = 7.5, t 0.66, and so f(0.66) = g(0.66) 40 The number of U.S. broadband Internet households will equal the number of dial-up Internet households (40 million each) around August of 004. Beyond that the former exceeds the latter.. We solve the system 4x + p = 59 5x 6p = Solving the first equation for p, we find p= x+. Substituting this value of p into the second equation, we have x 6( x+ ) = 4 5x + 8x 8 = 4 x = 04 x = 8. Substituting this value of x into the equation 4 59 p= x we have p = (8) + = = 9 Thus, the equilibrium quantity is 8000 units and the equilibrium price is $9.. We solve the system x + 7p = 56 x p = 45 Solving the first equation for x, we obtain x = 7p or x = p+ 8. Substituting this value of x into the second equation, we obtain Straight Lines and Linear Functions 5

9 7 ( p+ 8) p= 45 p+ 84 p= 45 4p = 58 or p = 6. 7 Then x = ( 6) + 8 = + 8 = 7. Therefore, the equilibrium quantity is 7000 units and the equilibrium price is $6.. We solve the system p = x + p = x +. Substituting the first equation into the second, we find x + = x + 5x = 0 and x =. Substituting this value of x into the first equation, we obtain p = () + = 8. Thus, the equilibrium quantity is 000 units and the equilibrium price is $8. 4. We solve the system p = 0.x + 6 p = 0.5x +.5. Equating the two equations, we have 0.x + 6 = 0.5x x = 4.5 x = 0. Substituting this value of x into the first equation gives p = 0.(0) + 6 and p =. Thus, the equilibrium quantity is 0,000 units, and the equilibrium price is $. 5. Let x denote the number of DVD players produced per week, and p denote the price of each DVD player. a. The slope of the demand curve is given by p 0 = = x Using the point-slope form of the equation of a line with the point (000, 485), we have p 485 = ( 5 x 000 ) p= 5 x or p = 0.08x b. From the given information, we know that the graph of the supply equation passes through the points (0, 00) and (500, 55). Therefore, the slope of the supply curve 5 Straight Lines and Linear Functions

10 is m = = = Using the point-slope form of the equation of a line with the point (0, 00), we find lthat p 00 = 0.09x p = 0.09x c. Equating the supply and demand equations, we have 0.08x + 75 = 0.09x x = 45 or x = 500. Then p = 0.08(500) + 75 = 55. We conclude that the equilibrium quantity is 500 and the equilibrium price is $ We solve the system x + 4p = 800 x 0p = 000. Solving the first equation for x, we obtain x = -4p Substituting this value of x into the second equation, we obtain -4p p = 000 4p = 800 or p = 75. Substituting this value of p into the first equation, we obtain x + 4(75) = 800 or x = 500. Thus, the equilibrium quantity is 500 and the equilibrium price is $ We solve the system x + p = 500 x p = 00. Solving the first equation for p, we obtain p = 500 x. Substituting this value of p into the second equation, we obtain x (500 x) = 00 x = 00 or x = 00. Next, p = 500 (00) = 600. Thus, the equilibrium quantity is 00 and the equilibrium price is $ Let x denote the number of espresso makers to be produced per month and let p denote the unit price of these espresso makers. Straight Lines and Linear Functions 54

11 a. The slope of the demand curve is given by p x = 0 40 = Using the point-slope form of the equation of a line with the point (50, 40), we have p 40 = ( 5 x 50 ) p= 5 x = 5 x b. The slope of the supply curve is given by p x = = = Using the point-slope form of the equation of a line with the point (750, 60), we have p 60 = ( 75 x 750 ) p= 75 x p= 75 x+ 50. c. Equating the demand equation and the supply equation, we have 5 x+ 50 = 75 x x = 00 or x = 875. Next, p = ( ) + 50 = 75. Thus, the equilibrium quantity is 875 espresso makers and the equilibrium price is $ a. We solve the system of equations p = cx + d and p = ax + b. Substituting the first into the second gives cx + d = ax + b (c a)x = b d b d or x = c a. Since a < 0 and c > 0, and b> d > 0, and c a 0, x is well-defined. Substituting this value of x into the second equation, we obtain p a b d ab ad + bc ab bc ad = ( ) + b = = () c a c a c a Therefore, the equilibrium quantity is b d and the equilibrium price is bc ad. c a c a b. If c is increased, the denominator in the expression for x increases and so x gets smaller. At the same time, the first term in the first equation () for p decreases 55 Straight Lines and Linear Functions

12 (since a is a negative number) and so p gets larger. This analysis shows that if the unit price for producing the product is increased then the equilibrium quantity decreases while the equilibrium price increases. c. If b is decreased, the numerator of the expression for x decreases while the denominator stays the same. Therefore x decreases. The expression for p also shows that p decreases. This analysis shows that if the (theoretical) upper bound for the unit price of a commodity is lowered, then both the equilibrium quantity and the equilibrium price drop. 0. The breakeven quantity is found by solving the equation C(x) = R(x), or cx + F = sx; that is, x = F sc - ( s c). Substituting this value of x into R(x) = sx gives the breakeven revenue as Rx ( ) = s( F - ) sf sc = sc -. Our analysis shows that for a breakeven operation, the breakeven quantity must be equal to the ratio of the fixed cost times the unit selling price and the difference between the unit selling price and unit cost of production.. True. Px ( ) = Rx ( ) Cx ( ) = sx ( cx+ F) = ( s cx ) F. Therefore, the firm is making a profit if Px ( ) = ( s cx ) F> 0, or x > sc F - ( s c).. True. In the typical linear demand curve, p drops as x increases, that is, the straight line has negative slope.. Solving the two equations simultaneously to find the point(s) of intersection of L and L, we obtain m x + b = m x + b (m m )x = b b () a. If m = m and b b, then there is no solution for () and in this case L and L do not intersect. b. If m m, then Equation () can be solved (uniquely) for x and this shows that L and L intersect at precisely one point. c. If m = m and b = b, then () is satisfied for all values of x and this shows that L and L intersect at infinitely many points. a 4. Rewrite the equations in the form y b x c a y b b x c = + and = +, b and think of these equations as the equations of the lines L and L, respectively. Using the results of Exercise 9, we see that a. the system has no solution if and Straight Lines and Linear Functions 56

13 a a c c only if =, or ab ab = 0, and ; b. the system has a unique b b b b solution if and only if ab ab 0. c. the system has a infinitely many solutions if and only if a b a b = 0 and c b c =, or c b b c = 0. b USING TECHNOLOGY EXERCISES.4, page 5. (0.6, 6.). (0.57, 6.87). (.86, 0.04) 4. (4.56, ) 5. (86.909, 45.99) 6. (.55,.548) 7. a. b. (548, 7,997) c. x-intercept is approximately a. (49, 60,58) b. 48 units 9. a. C( x) = x; C( x) = x b. c. (50, 6) d. If the distance driven is less than or equal to 50 mi, rent from Acme Truck 57 Straight Lines and Linear Functions

14 Leasing; if the distance driven is more than 50 mi, rent from Ace Truck Leasing. 0. a. Randolph Bank: D ( t ) = t; Madison Bank: D ( t ) = t b. c. Yes; 4 years from now. a. p= x+ 84; p= x+ 60. a b. (90, 9) c. 90/wk; $9/radio b. 558 units; $75.5 Straight Lines and Linear Functions 58