(Solution Manual) Calculus For Business, Economics, And The Social And Life Sciences

Laurence,Smith,Gerald,Mckenna, 10

Math & Statistics

Solution Manual

10th Edition

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Solution Manual for Calculus For Business, Economics, And The Social And Life Sciences, The downloadable file consists of 387 pages, and the total size of the file is 2.58 MB. Purchase now only for highly discounted price.

Author(s): Laurence, Smith, Gerald, Mckenna

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Student Solutions Manual to accompany Calculus for Business, Economics, and the Social and Life Sciences Tenth Edition, Brief Laurence D. Hoffman Smith Barney Gerald L. Bradley Claremon McKenna College Prepared by Devilyna Nichols Purdue University[to be supplied by publisher]CONTENTS Chapter 1 Functions, Graphs, and Limits 1 1.1 Functions 11.2 The Graph of a Function 61.3 Linear Functions 141.4 Functional Models 191.5 Limits 261.6 One-Sided Limits and Continuity 30 Checkup for Chapter 1 33Review Problems 36 Chapter 2 Differentiation: Basic Concepts 43 2.1 The Derivative 432.2 Techniques of Differentiation 522.3 Product and Quotient Rules; Higher-Order Derivatives 572.4 The Chain Rule 642.5 Marginal Analysis; Approximations Using Increments 722.6 Implicit Differentiation and Related Rates 75 Checkup for Chapter 2 82Review Problems 84 Chapter 3 Additional Applications of the Derivative 93 3.1 Increasing and Decreasing Functions; Relative Extrema 933.2 Concavity and Points of Inflection 1033.3 Curve Sketching 1143.4 Optimization 1243.5 Additional Applied Optimization 132 Checkup for Chapter 3 141Review Problems 148 Chapter 4 Exponential and Logarithmic Functions 159 4.1 Exponential Functions 1594.2 Logarithmic Functions 1654.3 Differentiation of Logarithmic and Exponential Functions 1734.4 Additional Exponential Models 182 Checkup for Chapter 4 199Review Problems 205 iiiiv Contents Chapter 5 Integration 219 5.1 Antidifferentiation; the Indefinite Integral 2195.2 Integration by Substitution 2265.3 The Definite Integral and the Fundamental Theorem of Calculus 2335.4 Applying Definite Integration: Area Between Curves and Average Value 2385.5 Additional Applications to Business and Economics 2455.6 Additional Applications to the Life and Social Sciences 252 Checkup for Chapter 5 259Review Problems 262 Chapter 6 Additional Topics in Integration 273 6.1 Integration by Parts; Integral Tables 2736.2 Introduction to Differential Equations 2846.3 Improper Integrals; Continuous Probability 2926.4 Numerical Integration 300 Checkup for Chapter 6 307Review Problems 312 Chapter 7 Calculus of Several Variables 325 7.1 Functions of Several Variables 3257.2 Partial Derivatives 3297.3 Optimizing Functions of Two Variables 3367.4 The Method of Least Squares 3467.5 Constrained Optimization: The Method of Lagrange Multipliers 3537.6 Double Integrals 362 Checkup for Chapter 7 371Review Problems 375Chapter 1 Functions, Graphs, and Limits 1.1 Functions 1. f( x)=3x+5, f(0)=3(0)+5=5 f(−1)=3(−1)+5=2 f(2)=3(2)+5=11 3. f( x)=3x2+5x−2, f(0)=3(0)2+5(0)−2=−2, f(−2)=3(−2)2+5(−2)−2=0, f(1)=3(1)2+5(1)−2=6. 5.g(x)=x+1 x, g(−1)=−1+1 −1=−2, g(1)=1+1 1=2, g(2)=2+1 2=5 2. 7. h(t)=/radicalbig t2+2t+4, h(2)=/radicalbig 22+2(2)+4=2√ 3, h(0)=/radicalbig 02+2(0)+4=2, h(−4)=/radicalbig (−4)2+2(−4)+4=2√ 39.f( t)=(2t−1)−3/2=1 (√ 2t−1)3, f(1)=1 [√2(1)−1]3=1, f(5)=1 [√2(5)−1]3=1 [√ 9]3=1 27, f(13)=1 [√2(13)−1]3=1 [√ 25]3=1 125. 11.f( x)=x−|x−2|, f(1)=1−|1−2|=1−|− 1|=1−1=0, f(2)=2−|2−2|=2−|0|=2, f(3)=3−|3−2|=3−|1|=3−1=2. 13.h(x)=/braceleftbigg −2x+4i fx≤1 x2+1i f x>1 h(3)=(3)2+1=10 h(1)=−2(1)+4=2 h(0)=−2(0)+4=4 h(−3)=−2(−3)+4=10 15.g(x)=x 1+x2. Since 1 +x2/negationslash=0 for any real number, the domain is the set of all real numbers. 17.f( t)=√ 1−t. Since negative numbers do not have real square roots, the domain is all real numbers such that1−t≥0, ort≤1. Therefore, the domain is not the set of all real numbers. 19.g(x)=x 2+5 x+2. 12 Chapter 1. Functions, Graphs, and Limits Since denominators cannot be 0, the domain consists of all real numbers such that x/negationslash=−2. 21.f( x)=√ 2x+6. Since negative numbers do not have real squareroots, the domain is all real numbers such that2x+6≥0,orx≥−3. 23.f( t)=t+2 √ 9−t2. Since negative numbers do not have real square roots and denominators cannot be zero, the domain is theset of all real numbers such that 9 −t 2>0,namely −3<t< 3. 25.f (u)=3u2+2u−6 and g(x)=x+2,so f( g( x) ) =f( x+2)=3(x+2)2+2(x+2)−6 =3x2+14x+10. 27.f (u)=(u−1)3+2u2andg(x)=x+1,s o f( g( x) ) =f( x+1) =[(x+1)−1]3+2(x+1)2 =x3+2x2+4x+2. 29.f (u)=1 u2andg(x)=x−1, so f( g( x) ) =f( x−1)=1 (x−1)2. 31.f (u)=√ u+1 andg(x)=x2−1,so f( g( x) ) =f( x2−1) =/radicalbig (x2−1)+1 =/radicalbig x2=|x|. 33. f( x)=4−5x f( x+h)−f( x) h=4−5(x+h)−(4−5x) h 4−5x−5h−4+5x h=−5h h=−535.f( x)=4x−x2 f( x+h)−f( x) h=4(x+h)−(x+h)2−(4x−x2) h =4x+4h−(x2+2xh+h2)−4x+x2 h =4x+4h−x2−2xh−h2−4x+x2 h =4h−2xh−h2 h=h(4−2x−h) h =4−2x−h 37.f( x)=x x+1 f( x+h)−f( x) h=x+h (x+h)+1−x x+1 h =x+h x+h+1−x x+1 h·(x+h+1)(x+1) (x+h+1)(x+1) =(x+h)(x+1)−x(x+h+1) h(x+1)(x+h+1) =x2+hx+x+h−x2−xh−x h(x+1)(x+h+1) =h h(x+1)(x+h+1)=1 (x+1)(x+h+1) 39.f( g( x) ) =f(1−3x)=√ 1−3x g(f( x))=g/parenleftbig√x/parenrightbig =1−3√x To solve√ 1−3x=1−3√x,square both sides, so 1−3x=1−6√x+9x −3x=−6√x+9x 6√x=12x √x=2x squaring both sides again, x=4x2 0=4x2−x 0=x(4x−1) x=0,x=1 41.1 Functions 3 Since squaring both sides can introduce extraneous solutions, need to check these values. /radicalbig 1−3(0)?=1−3√ 0 1=1/radicalBigg 1−3/parenleftbigg1 4/parenrightbigg ?=1−3/radicalbigg 1 4 1 2?=1−3 2 1 2/negationslash=−1 2 Also check remaining value to see if is in domain offandgfunctions. Since f(0)andg(0)are both defined, f( g( x) ) =g(f(x)) whenx=0. 41. f(g(x))=f/parenleftbiggx+3 x−2/parenrightbigg =2/parenleftBig x+3 x−2/parenrightBig +3 x+3 x−2−1=x g(f( x))=g/parenleftbigg2x+3 x−1/parenrightbigg =2x+3 x−1+3 2x+3 x−1−2=x Answer will be all real #’s for which fandgare defined. So, f(g(x))=g(f( x))for all real #’s except x=1 andx=2. 43. f( x)=2x2−3x+1, f( x−2)=2(x−2)2−3(x−2)+1 =2x2−11x+15. 45. f( x)=(x+1)5−3x2, f( x−1)=[(x−1)+1]5−3(x−1)2 =x5−3x2+6x−3. 47.f( x)=√x, f( x2+3x−1)=√ x2+3x−1. 49. f( x)=x−1 x, f( x+1)=(x+1)−1 x+1 =x x+1.51.f( x)=(x−1)2+2(x−1)+3 can be rewritten as g(h(x)) withg(u)=u2+2u+3 andh(x)=x−1. 53.f( x)=1 x2+1 can be rewritten as g(h(x)) with g(u)=1 u and h(x)=x2+1. 55.f( x)=3√ 2−x+4 2−x can be rewritten as g(h(x))with g(u)=3√u+4 u and h(x)=2−x. 57.D(x)=−0.02x+29;C(x)=1.43x2+18.3x+ 15.6 (a)R(x)=xD(x) =x(−0.02x+29) =−0.02x2+29x P(x)=R(x)−C(x) =/parenleftBig −0.02x2+29x/parenrightBig −/parenleftBig 1.43x2+18.3x+15.6/parenrightBig =−1.45x2+10.7x−15.6 (b)P(x)> 0 when −1.45x2+10.7x−15.6>0 Using the quadratic formula, the zeros of P are x=−10.7±/radicalbig (10.7)2−(4)(−1.45)(−15.6) 2(−1.45) x=2,5.38 so,P(x)> 0 when 2 <x< 5.38. 59.D(x)=−0.5x+39;C(x)=1.5x2+9.2x+67 (a)R(x)=xD(x) =x(−0.5x+39) =−0.5x2+39x P(x)=R(x)−C(x) =/parenleftBig −0.5x2+39x/parenrightBig −/parenleftBig 1.5x2+9.2x+67/parenrightBig =−2x2+29.8x−674 Chapter 1. Functions, Graphs, and Limits (b)P(x)> 0 when −2x2+29.8x−67>0 Using the quadratic formula, the zeros of P are x=−29.8±/radicalbig (29.8)2−(4)(−2)(−67) 2(−2) x≈2.76,12.14 so,P(x)> 0 when 2 .76<x< 12.14. 61.C(q)=0.01q2+0.9q+2 (a)C(10)=0.01(10)2+0.9(10)+2 =$12 (b)Cost of 10th unit =Cost of 10 units −Cost of 9 units =C(10)−C(9) C(9)=0.01(9)2+0.9(9)+2=$10.91 =$12−$10.91=$1.09 63.W(x) =600x 300−x (a)300−x/negationslash=0 x/negationslash=300 The domain is all real numbers except 300. (b)Typically, the domain would be restricted to theh first quadrant. That is, x≥0. However, sincexis a percentage, the restriction should be 0≤x≤100. (c)When x=50, W(50)=600(50) 300−50 =120 worker-hours (d)To distribute to all of the households, x=100 and W(100)=600(100) 300−100 =300 worker-hours(e)Need to find xwhenW(x) =150. 150=600x 300−x (150)(300−x)=(1)(600x) 300−x=4x x=60 After 150 worker-hours, 60% of the households have received a new telephone book. 65.C(x)=150x 200−x (a)All real numbers except x=200. (b)All real numbers for which 0 ≤x≤100. If x<0o rx>200 then C(x) < 0 but cost is non-negative. x>100 means more than 100%. (c)C(50)=150(50) 200−50=50 million dollars . (d)C(100)=150(100) 200−100=150 C(100)−C(50)=100 million dollars. (e)150x 200−x=37.5 187.5x=37.5(200), x=7,500 187.5=40%. 67.P(t)=20−6 t+1 (a)P(9)=20−6 9+1or 19,400 people. (b)P(8)=20−6 8+1 P(9)−P(8)=20−3 5−/parenleftBig 20−2 3/parenrightBig =1 15 This accounts for about1 15of 1,000 people, or 67 people. (c)P(t) approaches 20, or 20,000 people. Writing exercise–Answers will vary. 69. S(r)=C(R2−r2) =1.76×105(1.22×10−4−r2).1.1 Functions 5 (a)S(0)=(1.76×105)(1.44×10−4)= 25.344cm /sec. (b)S(0.6×10−2) =1.76×105(1.44×104−0.62×10−4) =1.76×105(1.08×104) =19.008 cm/sec . 71.s(A)=2.93√ A (a)s(8)=2.93√ 8=2.9×2=5.8 Since the number of species should be aninteger, you would expect to find approximately6 species. (b)s 1=2.93√ Aands2=2.93√ 2A s2=2.93√ 23√ A=3√ 2/parenleftBig 2.93√ A/parenrightBig =3√ 2s1. (c)100=2.93√ A 100 2.9=3√ A /parenleftbigg100 2.9/parenrightbigg3 =/parenleftBig 3√ A/parenrightBig3 /parenleftbigg100 2.9/parenrightbigg3 =A Need an area of approximately 41,002 square miles. 73.Q(p)=4,374 p2and p(t)=0.04t2+0.2t+12 (a)Q(t)=4,374 (0.04t2+0.2t+12)2 (b)Q(10)=4,374 (4+2+12)2=4,374 324 =13.5 kg/week . (c)30.375=4,374 (0.04t2+0.2t+12)2 (0.04t2+0.2t+12)2=4,374 30.375=144=122 So 0.04t2+0.2t+12=±12.The positive root leads to t(0.04t+0.2)=0 ort=0.(Disregard t<0.)The negative root produces imaginary numbers. t=0n o w . 75. (a) c(p)=0.4p+1 andp(t)=8+0.2t2 c(t)=0.4(8+0.2t2)+1=0.08t2+4.2 PPM. (b)c(2)=0.08(2)2+4.2=0.32+4.2= 4.52PPM . (c)6.2=0.08t2+4.2,t2=2 0.08=25,ort=5 years. 77.To find the domain of f( x)=4x2−3 2x2+x−3, Press y=. Enter(4x∧2−3)÷(2x∧2+x−3)fory1= Press graph. For a better view of the vertical asymptotes, press zoom and enter Zoom ln. Use arrow buttons to move cross-hair to the left-most vertical asymptote. Whenit appears cross-hair is on the line, zoom in againfor a more accurate reading. Move cross-hair againto be on the line. It appears that x=−1.5 is not in the domain of f. Zoom out once to move cross-hair to the rightmost vertical asymptote and repeat theprocedure of zoom in to find that x=1 is not the domain of f. The domain consists of all values except x=−1.5 andx=1. 79.Forf( x)=2√ x−1 andg(x)=x3−1.2,to find f(g(2.3)),we must find g(2.3) first and then input that answer into f. Press y=. Input 2√(x−1)fory1=and press enter. Inputx∧3−1.2 fory2=. Use the window dimensions [ −15,15]1 by [−10,10]1. Use the value function under the calc menu, input 2.3, and press enter. Use↑and↓arrows to be sure that y2=x∧3−1.2 is displayed in the upper left corner. The lower rightcorner display should read y=10.967 Use the value function again and input 10.967.Verify y 1=2√(x−1)is displayed in the upper left corner.The answer of y=6.31 is displayed in lower right corner.6 Chapter 1. Functions, Graphs, and Limits 81. (a) VE1 Level of Year Education 1991 1992 No H.S.diploma16,582 16,582=11 H.S. diploma24,007 16,582=1.4523,908 16,344=1.46 Some college27,017 16,582=1.6326,626 16,344=1.63 Bachelor’s degree41,178 16,582=2.4841,634 16,344=2.55 Advanced degree60,525 16,582=3.6562,080 16,344=3.80 1993 1994 1995 111 24,072 15,889=1.5224,458 16,545=1.4825,180 16,465=1.53 26,696 15,889=1.6826,847 16,545=1.6228,037 16,465=1.70 43,529 15,889=2.7444,963 16,545=2.7243,450 16,465=2.64 69,145 15,889=4.3567,770 16,545=4.1066,581 16,465=4.04 1996 1997 1998 111 25,289 17,135=1.4825,537 17,985=1.4225,937 17,647=1.47 28,744 17,135=1.6829,263 17,985=1.6330,304 17,647=1.72 43,505 17,135=2.5445,150 17,985=2.5148,131 17,647=2.73 69,993 17,135=4.0870,527 17,985=3.9269,777 17,647=3.95 1999 2000 11 26,439 17,346=1.5227,097 18,727=1.45 30,561 17,346=1.7631,212 18,727=1.67 49,149 17,346=2.8351,653 18,727=2.76 72,841 17,346=4.2072,175 18,727=3.85(b)1.45, 1.67, 2.76, 3.85. Writing exercise—Answers will vary. 1.2 The Graph of a Function 1.Since x-coordinate is positive and y-coordinate is positive, point is in quadrant I. (4, 3) xy 3.Since x-coordinate is positive and y-coordinate is negative, point is in quadrant IV . (5, –1)xy 5.Since x-coordinate is zero and y-coordinate is negative, point is on y-axis, below the x-axis. (0, –2)xy1.2 The Graph of a Function 7 7.P(3,−1),Q(7,1) D=/radicalBig (x2−x1)2+(y2−y1)2 =/radicalbig (1−(−1))2+(7−3)2 =√ 4+16=√ 20=√ 4·5=2√ 5 9.P(7,−3),Q(5,3) D=/radicalBig (x2−x1)2+(y2−y1)2 =/radicalbig (3−(−3))2+(5−7)2 =√ 36+4=√ 40=√ 4·10=2√ 10 11. (a) Since of form xn, where nis non-integer real number, is a power function. (b)Since of form anxn+an−1xn−1+...+a1x+ a0,where nis nonnegative integer, is polynomial function. (c)Since can multiply out and simplify to formanxxn+an−1xn−1+...+a1x+a0, is polynomial function. (d)Since is quotient of two polynomial functions is a rational function. 13.f( x)=x A function of the form y=f( x)=ax+b is a linear function, and its graph is a line. Two points are sufficient to draw that line. The x-intercept is 0, as is the y-intercept, and f(1)=1. (0, 0)(1, 1) xy15.f( x)=√x A function of the form y=√xis the positive half of the function y2=x(a parabola with vertex (0,0), a horizontal axis and opening to the right). The x- intercept and y-intercept are the same, namely (0,0). Choosing two more points on y=√x(for example P(1,1)andQ(2,4)), helps outline the shape of the half-parabola. (0, 0)(1, 1) 12.54 1 2.5 4xy 17.f( x)=2x−1 A function of the form y=f( x)=ax+bis a linear function, and its graph is a line. Two points are sufficient to draw that line. The x-intercept is1 2and they-intercept is −1. (0, –1)( , 0)1/H500821 /H1100222 /H1100211 /H1100212 /H1100220xy 19.Since function is of form y=Ax2+Bx+C (where C=0), the graph is a parabola; its vertex is/parenleftbigg −5 4,−25 8/parenrightbigg ,it opens up (A is positive), and its intercepts are (0,0) and/parenleftbigg −5 2,0/parenrightbigg .8 Chapter 1. Functions, Graphs, and Limits (0, 0)( , 0) 5/H11002/H5008 2 /H1100255 5 /H1100250y x 21.Since function is of form y=Ax2+Bx+C,the graph is a parabola which opens down (A is negative)and its vertex is (−1,16). Further, f( x)=−x 2−2x+15 =−(x2+2x−15) =−(x+5)(x−3). So the x-intercepts are (−5,0)and(3,0),and the y-intercept is (0,15). (0, 15) (–5, 0) (3, 0)xy 23.f( x)=x3 Note that if x>0 then f( x)> 0 and if x<0, then f( x)< 0.This means that the curve will only appear in the first and third quadrants. Since x3and(−x)3 have the same absolute value, only their signs are opposites, the curve will be symmetric with respectto (wrt) the origin. The x-intercept is 0, as is the y-intercept.(0, 0)(2, 8) xy 25.f( x)=/braceleftbigg x−1i fx≤0 x+1i fx>0 Note that the graph consists of two half lines on either side of x=0. There is no x-intercept for either half line. The half line y=x−1 has a y-intercept of −1,while the half line y=x+1 has no y-intercept. (0, 1) (/H110021, /H110022)(0, /H110021)(1, 2) xy 27.Graph consists of part of parabola y=x2+x−3, namely portion corresponding to x<1,and a half line for x≥1; for the parabola portion of the graph, the vertex is/parenleftbigg −1 2,−13 4/parenrightbigg ,and the parabola opens up (A is positive);/parenleftBigg −1−√ 13 2,0/parenrightBigg and(0,−3) are its intercepts; the half line starts at (1 ,−1) and includes the point (2 ,−3).1.2 The Graph of a Function 9 ( , 0) (1, /H110021) /H110021 /H11002 √13/H5008/H5008/H5008/H5008/H5008/H5008/H5008 2 /H1100255 50xy 29.y=3x+5 andy=−x+3 Add 3 times the second equation to the first. Then 4y=14 or y=7 2.Substitute in the first, then x=3−y=−1 2.The point of intersection is P/parenleftbigg −1 2,7 2/parenrightbigg . (– , 0)5 3(– , )1272(0, 5) (0, 3) (3, 0) xy 31.y=x2andy=3x−2 Setting the expressions equal to each other, x2=3x−2 x2−3x+2=0 (x−1)(x−2)=0 x=1,2 So points of intersection are P1(1,1)andP2(2,4).(1, 1)(2, 4)6 1 3 /H110023 /H1100240xy 33.3y−2x=5 andy+3x=9. Multiply the second equation by −3 and add it to the first one. Then, −2x−9x=5−27, x=2,y=9−3(2)=3. The point of intersection is P(2,3). (2, 3) /H1100252.5 5 /H110025010 xy 35. (a) Crosses y-axis at y=−1,y-intercept is (0,−1). (b)Crosses x-axis at x=1,x-intercept is (1 ,0). (c)Largest value of fis 3 and occurs at x=4 (highest point on graph). (d)Smallest value of fis−3 and occurs at x=−2 (lowest point on graph). 37. (a) Crosses y-axis at y=2,y-intercept is (0 ,2). (b)Crosses x-axis at x=−1 and 3.5; x-intercepts are(−1,0)and (3.5,0).10 Chapter 1. Functions, Graphs, and Limits (c)Largest value of fis 3 and occurs at x=2 (highest point on graph). (d)Smallest value of fis−3 and occurs at x=4 (lowest point on graph). 39.The monthly profit is P(p)=(number of recorders sold) (price−cost) =(120−p)(p−40) So, the intercepts are (40 ,0), (120 ,0), and (0,−4800). The graph suggests a maximum profit whenp≈80, that is, when 80 recorders are sold. P(80)=(120−80)(80−40)=1600 So estimated max profit is $1600. pP (80, 1,600) 120 40 41.The weekly profit is P(x)=(number of sets sold ) ·(price-cost per set ) =5(27−x)(x−15) So, the intercepts are (27,0),(15,0)and(0,−2025). The graph suggest a maximum weekly profit whenx≈21. That is, when the price per set is $21. P(21)=5(27−21)(21−15)=180 So, estimated max profit is $180(21, 180) 15 30xP The number of sets corresponding to the max profit is S(27−21)=30 sets 43. (a) E(p)=(price per unit)(demand) =−200p(p−60) (30, 180000) 2 /H11003 105 1 /H11003 105 60 03 0pE (b)Thepintercepts represent prices at which consumers do not buy commodity. (c)The graph suggests a maximum expenditure whenp≈30. E(30)=−200(30)(30−60)=180,000 So estimated max expenditure is $180,000. 45.H(t)=−16t2+160t=−16t(t−10) (a)The intercepts of the graph are (0,0)and(10,0). Due to symmetry, the vertex is when t=5 and y=H(5)=−16(5)2+160(5)=400.1.2 The Graph of a Function 11 (5, 400) 5100200300400 10tH (b)Aside from when it is initially thrown, the height of the projectile is zero (ground level) whent=10 seconds. (c)The high point of the graph, which corresponds to the max height, is when y-coordinate is 400, or 400 feet. 47. (a) profit=revenue −cost =(#sold)(sellingprice )−cost P(x)=x(−0.05x+38)−(0.02x 2+3x+574.77) =−0.07x2+35x−574.77 hundred dollars (250, 3800.23)4,000 2,000 500 0 250x (100)P ($100) (b)The graph suggests a maximum profit when x=250, that is, when 25,000 units are purchased. Note that the max profit is P(250)= −0.07(250)2+35(20)−574.77≈3800.23 hundred, or $380,023. For the unit price, p=−0.05(250)+38=$25.5049.D(v)=0.065v2+0.148v For practical domain, graph is part of parabolacorresponding to v≥0. 700 350 100 05 0vD 51. (a) revenue =(#apts) (rent per apt) Sincep−1200 100represents the number of $100 increases, 150−5/parenleftbiggp−1200 100/parenrightbigg =210−0.05p represents the number of apartments that will be leased. So, R(p)=210p−0.05p2 (b) (2,100, 220, 500)250,000 125,000 4,000 2,000pR (c)The graph suggests a maximum profit when p=2100; that is, when the rental price is $2,100. The max profit is R(2100)= 210(2100)−0.05(2100)2≈$220,500.