Chapter 1: Linear Equations and Functions

Transcription

1 Chapter : Answers to Eercises Chapter : Linear Equations and Functions Eercise.. 7= = = = 8( + ). + 8= = = 0 = 0 9 = = = () = 96 = 7. ( 7) = ( + ) 9.. = + + = + = 9 = 9 7 = = 6 6 = 7 = 7 = 7 7 = + = 6 + = 6 + = = = = =.. 7. ( ) = () = = = =? Check: = () 0? = = = is the solution. = + (+ ) Multipl each term b 6( + ). = (8+ 0) 8 = 0 = or = ( )? Check: = + + ( ) ( )? = + 0 = and = 0 = is the solution. + = = 6 ( ) + = There is no solution.

2 Chapter : Answers to Eercises =.8( ) = = = = =.( ) = = = = = = + = + =. 9+ = () 8+ = = 8+ = S = P+ Prt Prt = S P S P t = Pr 9. ( ) < < < <. > 9 > 8 ( )( ) > 8( ) <. ( ) ( ) (. ( ) > + > + > + > > 6 ( )( ) > 6( ) ) < > > ( ) > 6 > 6 ( )( ) > 6( ) < ( )( ) < ( )( ) ( ) < 6 9 < 8 ( ) 9 < < + 8 < 8 < < = 68, , 000 = 68, = 68, , 000 = 6, 000 6,000 = = months 800

3 Chapter : Answers to Eercises. I = r 7.9 I = I = = I = I = $6.0. R = C for breakeven point 0 = = 790 ( ) = 0 packs or 0,000 CD's 7. 70,00 = , 00 = = $9, C 00I = 6.8 ( ) 99C 00 7 = C 7000 = C = 76.8 C = 78.68% FE + FE = 90 6 FE + 8 = 0 FE = 9 FE = 96 A 96 is the lowest grade that can be earned on the final.. = amount in safe fund,000 = amount in risk fund Yield: (,000 ) =, , =, = 600 = 90,000 = $90,000 in 9% fund, = $0,000 in % fund.. Reduced salar: (000) = $800 Increased salar: (800) = $60 60 = R% of R = = $60 is an 8% increase > > 600 > t 900.7t 0 t.6 He could bu tapes. 6. a = b t > 0 c. 008.t >.79 t >.8 6. A = h a h.h 9.8 h 0.8 b h < 0.h < 9.8 h < 0. Eercise.. a. For each value of there is onl one. b. D = { 7,,0,,.,9,,,8, } R = { 0,,,9,,,,60}. f(0) =, f() =. This is a function, since for each there is onl,,,8, 9 R =,,6 one. D = { }, { } 7. The vertical line test shows that graph (a) is a function of, and that graph (b) is not a function of.

4 Chapter : Answers to Eercises 9. If =, then is a function of.. If =, then is not a function. If, for eample, =, there are two possible values for.. R() = 8 a. R(0) = 8(0) = b. R() = 8() = 6 c. R( ) = 8( ) = d. R(.6) = 8(.6) = C a. b. c. d. ( ) = f( ) a. b. c. C (0) = (0) = C( ) = ( ) = C( ) = ( ) = C = = 6 = f = = + 8 = 8 f () = = 8 = 6 f ( ) = ( ) = 8 + = 6 f ( ) = + + a. f ( + ) = f() = + + = f() + f() = 7 + = f() + f() f( + ) b. f ( + h) = + ( + h) + ( + h) c. No. This is equivalent to a. d. f ( ) + h = h e. No. f( + h) f( ) + h 6 8 f( + h) = + ( + h) + ( + h) = + + h+ + h+ h f( ) = + + f( + h) f( ) = h+ h+ h = h(+ + h) f( + h) f( ) = + + h h. f ( ) = a. b. f( + h) = ( + h) ( + h) = + + f( + h) f( ) h h h = + + ( h) ( h) ( ) = + h h h + = h h h f( + h) f( ) h h h = h h = h. Since (9, ) and (, 6) are points on the graph: a. f (9) = b. f () = 6. a. The ordered pair (a, b) satisfies the equation. Thus b= a a. b. The coordinates of Q = (, ). Since the point is on the curve, the coordinates satisf the equation. c. The coordinates of R = (, ). The satisf the equation. d. The values are 0 and. These values are also solutions of = = + There is no division b zero or square roots. Domain is all the reals, i.e., { : Reals}. Since 0, +, the range is reals or { : }. 9. = + There is no division b zero. To get a real number, we must have + 0 or. Domain:. The square root is alwas nonnegative. Thus, the range is : reals, 0.. D: { :, }. D: { : 7 7}. f( ) =, a. b. c. g( ) { } = ( f + g)( ) = + ( f g)( ) = ( f g)( ) = = f g d. ( ) = =

5 Chapter : Answers to Eercises 7. f ( ) =, a. b. c. g( ) = ( f + g)( ) = + ( f g)( ) = ( )( ) = = f g f d. ( ) = g 9. f( ) = ( ), g( ) = a. b. c. d. ( f g)( ) = f( ) = ( ) = 8 ( g f)( ) = g(( ) ) = ( ) f( f( )) = f(( ) ) = [( ) ] 6 ( f f)( ) = ( ) ( ) = ( ). f ( ) =, a. b. ( f f )( ) f ( f( ) ) g ( ) = + ( f g)( ) f( ) = + = + = = + ( g f)( ) g( ) ( ) = + 6 c. f ( f( )) = f( ) = d. ( f f)( ) = = ( f f )( ) f ( f( ) ). a. f (0) =, 000 means it will take 0 ears to pa off a debt of $,000 (at $800 per month and 7.% compounded monthl.) b. f( + ) = f() = 69, 000 ; f() + f() = 80, 000 ; No. c. It will take ears to pa off the debt, i.e., 89,000 = f ().. a. f (90) = 6. means that in 90 there were 6. workers supporting each person receiving Social Securit benefits. b. f (990) =. c. The points based on known data must be the same and those based on projections might be the same. d. Domain: 90 t 00 Range:.9 n a. f(9) =,000,000, g(9) = 600,000 b. f (0) =, 00,000. This was the number of prisoners in 000. c. g (90) = 00, 000. This was the number of parolees in 990. d. ( f g)(0) =,00, ,000 = 00,000. There were this man more in prison than were out on parole. f g (9) = 900, ,000 = 00,000 e. ( ) ( f g) (98) =,0, ,000 = 00,000 ( f g)(98) is greater. Possible reason is increased prison capacit but parolee level is constant. 9. a. Since the wind speed cannot be negative, s 0... b. f () = () 9.6 = 9. At a temperature of F and a wind speed of mph, the temperature feels like 9. F. c. f (0) =.69, but f (0) should equal the air temperature, F. C a. ( ) = () 00() 0.() 0 C = + + = (0) + 0 = = $ b. The value C(0) is the total cost of producing 0 units. c. C (0) = 00(0) + 0.(0) p C( p) = 0 p = $, 00 a. Domain: { p: 0 p< 0} b. c. d. e. 700() 8,00 C () = = = $ (90) 67,000 C (90) = = = $6, (99) 7,700 C (99) = = = $7, (99.6) 77,080 C (99.6) = = = $,87, In each case above, to remove p% of the particulate pollution would cost C( p).

6 6 Chapter : Answers to Eercises. a. A is a function of. b. A() = (0 ) = 96 sq ft A(0) = 0(0 0) = 600 sq ft c. For the problem to have meaning we have 0 < < a. Pqt ( ( )) = P(00+ t) (00 + t) = 80(00 + t) 00 0 = 69, t t b. q() = 00 + () = 0 P(q()) = $9,7 9. R = f( C) C = g( A) a. ( )( ) ( ) f g = f C = R b. ( g f )( ) is not defined. c. A is the independent variable and R is the dependent variable. Revenue depends on mone spent for advertising. 6. length = width = L = = or = L = + = + 6. Revenue = (no. of people)(price per person) Eample: R = 0 R = 9.80 R = 9.60 Solution: R = (0 + )( 0.0) Eercise.. + = -intercept: = 0 then =. -intercept: = 0 then =.. + = 0 -intercept: = 0 then = 0 Likewise, -intercept is = 0. + = -. 8 = 60 -intercept: = 0 then -intercept: = 0 then = 60 =. = (, ) and (, 7) 7 8 m = = = 7 = 9. (, ) and (, ) 0 m = = = = 0 ( ). A horizontal line has a slope of 0.. (, ) and (, ) The rate of change is equivalent to the slope of the line. 0 m = = = = 0. a. Slope is negative. b. Slope is undefined.

7 Chapter : Answers to Eercises =, m =, b = = 6. m =, b = = = or = 0+, m = 0, b = 6 = - - = m =, b = = +. = 8 Slope is undefined. There is no -intercept. = 8 - = P(, 0), m = ( ) 0= = +. + = 6 or = +, m =, b =. 6 + = =

8 8 Chapter : Answers to Eercises,, m = = ( ( ) ) 9 = +. P( ) P(, ), m is undefined = =. P = (, ), P = (, 6) 6 m = = = ( ) = 9 = = 6 = 6 = + = Lines are perpendicular since =.. 6 = 6 = or = Lines are the same. = 6 =. If + =, then = +. So, m =. A line parallel will have the same slope. Thus, m = and P = (, 7) gives ( 7) = ( ( )) which simplifies to =.. If 6 =, then =. Slope of the perpendicular line is. Thus m = and 6 P = (, ) gives = ( ) which 6 simplifies to = P = (7, ), P = ( 6, ) m = = = 6 7 = ( 7) 7 = + = + or + =

9 Chapter : Answers to Eercises 9 7. a. 60,000 80,000 = 60, b. 0 = 60, = = 0 months 00 In 0 months, the building will be completel depreciated. c. (60, 70,000) means that after 60 months the value of the building will be $70, =.7+.6 a. m =.7, b =.6 b. In 99, when = 0,.6% of the U.S. population had Internet. c. The percentage of the U.S. population with Internet is changing at the rate of.7% per ear.. M( ) = a. m= 0.76 b= 8.8 b. In 90, 8% of the unmarried women became married. c. The annual rate of change is 0.76%. For each passing ear the percent of unmarried women who get married decreases b 0.76%.. F = 0.M a. m = 0. b. For each $ increase in male earnings, the female s earning increases onl b $0.. F 0 = c. ( ) ( ) = 8.80 thousands =$8,80. = Both units are in dollars B 7 = a. m = = 0.69 and ( 676, 7) ( W ) B = 0.690W B (80) = b. ( ) = $ a. ( 980,8. ) and ( 00,79.9 ) m = = = =. 980 ( ) = b. The consumer price inde for urban consumers increases at the rate of $. per ear. 6. (, p) is the reference. (0, 8000) is one point. 700 m = = 700 p 8,000 = 700( 0) or p = , (t, R) is the ordered pair. 7 P =,, P = (6,9) m = = = =. 7 6 R 9 =.( t 6) or R =.t or R =.t P = (00, ) P = (0, 9) 9 m = = = = 0.8( 00) or = 0.8 7

10 0 Chapter : Answers to Eercises Eercise.. = + +. = a. = = b. 000 = a. + = = = + b. Standard Window + = = 0 7. a. -intercept = 0.0 -intercept: 0.00 = 0.0 b. 0. = 0 = =

11 Chapter : Answers to Eercises 9. Complete graphs can be seen with different windows. A hint is to look at the equation and tr to determine the ma and/or min of. Also, find the - intercepts. 9. ( ) = There is no min. Ma value of =. -intercepts: 0 0.(.) = = 0. = + (.) = ± ± 8 =. ± 8 or =. or 8. ( ) = If = 0, =. A suggested window is shown below. = = 6 6 = + = + f ( ) = + ( ) ( ) = + f ( ) = + = 8 + = 8 f = + = 0.77 Use our graphing calculator, and evaluate the function at these two points. If either of our answers differ, can ou eplain the difference?. As gets large, approaches. When = 0, =, intercepts at ±.. = + 6 ( )( + ) = = ( + )( + ) What happens to as approaches??. = 8 = = = = + = + = 8. 6 = = 7 = = 6 = 0 ( )( ) + = 0 = or = + 9. Find the zeros and find the -intercepts are equivalent statements. Use a graphing calculator s TRACE or ZERO. a.-b. Graphing calculator approimation is =.98, 8.98.

12 Chapter : Answers to Eercises. a. 0 F = 0. M b. If males average $0,000, then females earn $,. F 6. =.99 thousand = $,00 c. ( ) d. p = a. E =,000p 0 p 00, a. e. The percentage decreases before 000 and increases after ,000 8, 000 C = 80 p 0 0, a. b. E 0 when 0 p 0. 0, R = 8, , 000 b. The rate is As more people become aware of the product, there are fewer to learn about it. 7. a. -min = 0, -ma = b. -min = 0, -ma = 70 c. p = a. b. Near p = 0, cost grows without bound. c. The coordinates of the point mean that the cost of obtaining stream water with % of the current pollution levels would cost $8,0. d. The p-intercept means that the cost of stream water with 0% of the current pollution levels would cost $0., = b. Increasing. The per capita federal ta burden is increasing. 0 0 Eercise.. Solution: (, ). Infinitel man solutions. : = + : + = : = : = +

13 Chapter : Answers to Eercises. Solution: (, ) 7. Solution: No solution since the graphs do not : = + : = intersect. : = : = 9.. = 6 = 8 Solve for. = = Substitute for this variable in ()=6 first equation and solve for = 6 + = the other variable. The solution of the sstem is = = 8 = and =, or,. + = 6 Solve for. = ( ) Substitute for this variable in + = 6 second equation and solve for = 6 the other variable. = 6 Solve for : = = 6 6 The solution of the sstem is = and =, or,.. + = Multipl st equation b. 9+ = = Multipl nd equation b. 8 = 8 Add the two equations. 7 = Solve for the variable. = Substitute for this variable in () + = either original equation and = 8 solve for the other variable. = The solution of the sstem is = and =, or (, ). = /. + = Multipl first equation b. + 9 = = Multipl second equation b. 8 = 6 Add the two equations. = Substitute for this variable in + () = either original equation and = 8 solve for the other variable. = The solution of the sstem is = and =.

14 Chapter : Answers to Eercises = =. =. Multipl nd equation b = 0.6 The solution of the sstem is Subtract the two equations. 0.9 =.6 Solve for the variable. Substitute, solve for. = = = and =, or, = Multipl first equation b 6. = = Multipl second equation b = 77 Add the two equations. 7 = 7 Substitute for this variable in = either original equation and 8() + = solve for the other variable. = The solution of the sstem is = and =, or (, ). =. + 6 = + 6 = + = Multipl second equation b. 6 = Add the two equations: 0 = 0 There are infinitel man solutions. The sstem is dependent. Solve for one of the variables in terms of the remaining variable: =. Then a general solution is c, c, where an value of c will give a particular solution. Use the standard window and graph each equation. Use the TRACE or INTERSECT feature to find the solution.. = 8 = = + 8. :+ = :+ 7 = = = = Solution: (, ) Solution: (, )

15 Chapter : Answers to Eercises Eq. + + z = Steps,, and of the sstematic Eq. + z = 8 procedure are completed. Eq. z = Step : z = From Eq. + () = 8 or = 7 From Eq. + (7) + = or = 7 The solution is = 7, = 7, z =. Eq. 8z = 0 Steps and of the sstematic Eq. + z = 8 procedure are completed. Eq. + z = Step : ( ) Eq added to Eq. gives z = or z =. From Eq. + ( ) = 8 or = From Eq. 8( ) = 0 or = The solution is =, =, z = or (,, ). Eq. + z = 9 Step is completed. Eq. + + z = Eq. + 8z = Step : + z = 9 Eq. Eq. + z = ( ) Eq. added to Eq. Eq. 6z = ( ) Eq. added to Eq. Step is also completed. Step : z = from Eq.. From Eq. + = or = 9 From Eq. + ( 9) = 9 or = The solution is =, = 9, z = or, 9,. f( ) = , h( ) = = = 7.9 ( ) f ( ) =. during 998 h. =. = $8.67 billion

16 6 Chapter : Answers to Eercises. a. + = 800 Total number of tickets b. 0 = revenue from $0 tickets c. 0 = revenue from $0 tickets d =, 000 Total Revenue e. Multipl equation from part (a) b = = 000 = 6000 = 600 Substitution into equation from part (a) gives = 0. Sell 0 of the $0 tickets and 600 of the $0 tickets. 7. = amount of safe investment. = amount of risk investment. + =,600 Total amount invested = 0,000 Income from investments The solution is the solution of the above sstem of equations. + =, = 00,000 () second equation 0.8 =, 00 Subtract equations = 68,000 Solve for or amount of risk investment. Substituting = 68,000 into one of the original equations we have + 68,000 =,600 or = $77,600. Solution: Put $77,600 in a safe investment and $68,000 in a risk investment. 9. = amount invested at %. = amount invested at %. + =,000 Total amount invested =,000 Investment income Solve the sstem of equation: + =, =,00 () second equation 0. = 0,000 Subtract equations = 0,000 Solve for Substituting into the first equation we have + 0,000 =,000 or =,000. Thus, $,000 is invested at % and $0,000 is invested at %.

17 Chapter : Answers to Eercises 7. A = ounces of substance A. B = ounces of substance B. A Required ratio = gives A B = 0. B Required nutrition is %A + %B = 0%. This gives A + B = 0. The % notation can be trouble. Be careful! Now we can solve the sstem. A B = 0 A+ B = 0 B = 0 Subtract first equation from second. 0 0 B = = 0 Substituting into the original equation gives A = 0 or A=. The solution is ounces of substance A and 6 ounces of substance B.. = population of species A. = population of species B. + =,600 units of first nutrient + = 9,60 units of second nutrient 8+ =,00 () first equation + = 9,60 =,70 Subtract = 0 Solve for Substituting = 0 into an original equation we have (0) + =,600. So, = 00. Solution is 0 of species A and 00 of species B.. = amount of 0% concentration. = amount of % concentration. + = amount of solution = 0.() concentration of medicine Solving this sstem of equations: + = + 0. = 7.7 () second equation 0.7 =. Subtract equations = Solve for Substituting into the first equation we have + = or = 7. The solution is cc of % concentration and 7 cc of 0% concentration. 7. = number of $0 tickets. = number of $0 tickets. + = 6,000 total number of tickets = 80,000 total revenue To solve the sstem of equations multipl Eq. b = 80, = 80,000 = 0,000 or =,000 Substituting into the first equation gives = Sell,000 tickets for $0 each and 6000 tickets for $0 each.

18 8 Chapter : Answers to Eercises 9. = amount of 0% solution to be added. 0.0 = concentration of nutrient in 0% solution. 0.0(0) = is the concentration of nutrient in % solution = 0.( + 0) 0.0+ = = 8 or = 80cc of 0% solution is needed.. = ounces of substance A, = ounces of substance B, and z = ounces of substance C. + + z = 0 Nutrition requirements = z = z Digestive restrictions Digestive restrictions Since both and are in terms of z, we can substitute in the first equation and solve for z. So, z + z + z = 0 or 0z = 0. So, z =. Now, since = z, we have =. Since = z, we have =. The solution is ounces of substance A, ounce of substance B, and ounces of substance C.. A = number of A tpe clients. B = number of B tpe clients. C = number of C tpe clients. A + B + C = 00 Total clients 00A + 00B + 00C = 0,000 Counseling costs 00A + 00B + 0C = 0,000 Food and shelter To find the solution we must solve the sstem of equations. Eq. A+ B+ C = 00 Eq. A+ B+ C = 00 Original equation divided b 0 Eq. A+ B+ C = 00 Original equation divided b 0 A+ B+ C = 00 Eq. Eq. B+ C = 00 ( ) Eq. added to Eq. Eq. B C = 00 ( ) Eq. added to Eq. A+ B+ C = 00 Eq. B+ C = 00 Eq. C = Eq. added to Eq C = = 00 Substituting C = 00 into Eq. gives B + 00 = 00 or B = 00. So, B = 0. Substituting C = 00 and B = 0 into Eq. gives A = 00. So, A = 00. Thus, the solution is 00 tpe A clients, 0 tpe B clients, and 00 tpe C clients.

19 Chapter : Answers to Eercises 9 Eercise.6. a. P ( ) = R ( ) C ( ) ( ) = = 7 00 P 00 = = $700 b. ( ) ( ). a. P ( ) = R ( ) C ( ) b. c. = 80 (+ 80) = 7 80 P (0) = 7(0) 80 = $70 The total costs are more than the revenue. P ( ) = 0 or 7 80 = 0 80 So, = = 0 units is the break-even 7 point.. C() = + 0 a. m =, C-intercept: 0 b. MC = means that each additional unit produced costs $. c. Slope = marginal cost. C-intercept =fied costs. d. $, $ ( MC = at ever point) 7. R = 7 a. m = 7 b. 7; each additional unit sold ields $7 in revenue. c. In each case, one more unit ields $7. 9. R() = 7, C() = + 0 a. P ( ) = 7 (+ 0) = 0 b. m = c. Marginal profit is. d. Each additional unit sold gives a profit of $. To maimize profit sell all that ou can produce. Note that this is not alwas true.. (, P) is the correct form. P = (00, 0) P = (0, 6000) m = = P 0 = 8( 00) or P = The marginal profit is 8.. a. TC = H b. TR = 60H c. P = R C ( H ) = 60H = H 6600 C 00 = d. ( ) ( ) = $,600 cost of 00 helmets ( 00) 60( 00) R = = $,000 revenue from 00 helmets ( 00) = ( 00) ( 00) P R C = $, 000, 600 = $600 loss from 00 helmets e. C (00) = (00) = $7,0 cost of 00 helmets R (00) = 60(00) = $8,000 revenue from 00 helmets P(00) = R(00) C(00) = 8, 000 7,0 = $900 profit from 00 helmets f. The marginal profit is $. Each additional helmet sold gives a profit of $.. a. The revenue function is the graph that passes through the origin. b. At a production of zero the fied costs are $000. c. From the graph, the break-even point is 00 units and $000 in revenue or costs d. Marginal cost = = Marginal revenue = = R() = C() = 8 = + 60 or 0 = 60 or =. Thus, necklaces must be sold to break even. 9. a. R() =, C() = b. R() = C() if = or = 600 or = 00. It takes 00 units to break even.. a. P ( ) = R ( ) C ( ) = (8+ 600) = 600 b. B setting P() = 0 we get = 00. Same as 9(b).

20 0 Chapter : Answers to Eercises. a. TC =.0 + b. TR = c. P = R C ( ) =.0 + =.0 d. Breakeven also means P = 0..0 = 0.0 =. a. R() =.90 b. P = (000, 0000) P = (800, ) 7. a. = 90 units to break even, 0, 000 7,880 m = = = , 000 =.90( 000) or =.90+ 0, 00 = C( ) c. From.90 = ,00 we have = 0 units to break even. $ BE TR P VC b. TR starts at the origin and intersects TC at the break-even (BE). FC is a horizontal line from the vertical intercept of TC. VC starts at the origin and is parallel to TC. 9. If price increases, then the demand for the product decreases.. a. If p = $0, then q = 600 (approimatel). b. If p = $0, then q = 00. c. There is a shortage since more is demanded. TC FC. Demand: p+ q = 00 (60) + q = 00 q = 80 q = 6 Suppl: p q = 60 q = q = 0 q = There will be a surplus of 9 units at a price of $ Remember that (q, p) is the correct form. P = (0, 900) P = (, 80) m = = = 0 7 Note: m < 0 for demand equations. p 900 = ( q 0) or p = q (q, p) is the correct form. P = (000,.0) P = (000,.00) m = = = Note: m > 0 for suppl equations. p = 0.000(q 000) or p = 0.000q a. The decreasing function is the demand curve. The increasing function is the suppl curve. b. Reading the graph, we have equilibrium at q = 0 and p =.. a. Reading the graph, at p = 0 we have 0 units supplied. b. Reading the graph, at p = 0 we have 0 units demanded. c. At p = 0 there is a shortage of 0 units.. B observing the graph in the figure, we see that a price below the equilibrium price results in a shortage.

21 Chapter : Answers to Eercises. q+ 8 = q+ Required condition. 7. q+ 68 = q+ 68 Multipl both sides b 6 to simplif. q = 0 q = 0 Substituting into one of the original equations gives Thus, the equilibrium point is (q, p) = (0, 8). q+ 0 = q+ 0 Required condition. 90 = 9q p = (0) + 8 = 8. q = Solve for q. Substituting q = into one of the original equations gives p = 80. Thus, the equilibrium point is (q, p) = (, 80) Demand: (80, 0) and (, 00) are two points. m = = 80 p p = m( q q) or p 00 = ( q ) or p = q Suppl: (60, 80) and (0, 70) are two points. m = = p p = m( q q) or p 80 = ( q 60) or p = q Now, set these two equations for p equal to each other and solve for q. 9 q+. = q+ 0 Required for equilibrium. 8 9q+ 700 = q+ 600 Multipl both sides b 8 to simplif. 9q = 900 q = 0 Substituting q = 0 into one of the original equations gives p =. Thus, the equilibrium point is (q, p) = (0, ).. a. Reading the graph, we have that the ta is $. b. From the graph, the original equilibrium was (0, 0). c. From the graph, the new equilibrium is (0, ). d. The supplier suffers because the increased price reduces the demand.. New suppl price: p = q = q + 68 q+ 68 = q+ 0 Required condition 9q = q = 8 Substituting q = 8 into one of the original equations gives p = 88. Thus, the new equilibrium point is (q, p) = (8, 88). q q. New suppl price: p = + + = = + 6 Required condition q q 0 0 q+ 00 = q+ 00 q = 00 q = 00 Thus, p = + = The new equilibrium point is (00, 0).

22 Chapter : Answers to Eercises q + 0 q Demand: p = Suppl: p = 60 q+ 0 q q+ 600 New suppl: p = + = + = = q+ 600 q Required condition q+ 600 = q+ 00 Multipl both sides b q = 600 q = 0 Thus, p = = The new equilibrium quantit is 0. The new equilibrium price is $.