Solution Manual for Mathematical Modeling, 4th Edition

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1. An automobile manufacturer makes a profit of $1,500 on the sale of a certain model.It is estimated that for every $100 of rebate, sales increase by 15%.(a) What amount of rebate will maximize profit? Use the five-step method, and model asa one-variable optimization problem.Step 1: Ask the question.Variables:r= rebate ($)s= number of cars soldP= profit ($)Assumptions:s= s_0 (1+0.15(r/100))P= (1500-r) ss>= 0 , 0<= r <=1500where the constant s_0 is the number of sales without any rebateObjective:Maximize P.Step 2: Select the modeling approach.We will model this problem as a one variable optimization problem. See text p. 6.Step 3: Formulate the model.Let x=r and y=P, and writey = f(x) = (1500-x) s_0 (1+0.15(x/100)).Our goal is to maximize f(x) over the interval [0, 1,500].Step 4: Solve the model.Compute f '(x) = 1500 s_0 (0.0015) + (-1) s_0 (1+0.15(x/100)) = 0 at x = 416.6667,f(x) = 1760.42 and since the graph of f(x) is a parabola we know this is the globalmaximum.Step 5: Answer the question.According to this model, the optimal policy is to offer a rebate of around $420, whichshould result in about a 17% increase in profits as compared to no rebate.(b) Compute the sensitivity of your answer to the 15% assumption. Consider both theamount of rebate and the resulting profit.Generalize the model from part (a) and lety = f(x) = (1500-x) s_0 (1+e x)

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where currently e = 0.0015. Compute thatf '(x) = 1500 s_0 e + (-1) s_0 (1+e x) = 0at x = 750-1/(2e). Then dx/de = 1/(2 e^2), and soS(x,e) = (dx/de) (e/x) = 1/(2*0.0015^2) (0.0015/416.6667) = 0.8so if the rebate is 10% more effective then we thought, then the optimal rebate will be 8%bigger. A similar computation yields S(y,e) = 0.38 so that if the rebate is 10% moreeffective than we thought, then our optimal profit will be 4% greater.(c) Suppose that rebates actually generate only a 10% increase in sales per $100. Whatis the effect? What if the response is somewhere between 10 and 15% per $100 ofrebate?Using the results of part (b), if e decreases by 33% to 0.0010 then we expect the optimalrebate to decrease by (0.8) 33% = 27% to around $310. We would also expect profits togo down by (0.38) 33% = 13% to around $1530. A direct computation (solve the entireproblem again using e=0.0010) yields a similar result: a 40% decrease in the optimalrebate (to $250) and an 11% decrease in profit (to 1562.50).(d) Under what circumstances would a rebate offer cause a reduction in profit?Using the sensitivity results of part (b), if every $100 of rebate results in a sales increaseof less than 8.3% then the optimal policy is to offer no rebate. To see this, note thatcurrently the optimal profit is 17% higher than with no rebate, and (0.38) 44.7% = 17%.The current rebate effectiveness is e = 0.0015 and so a 44.7% decrease yields 0.00083.Exact calculations yield a similar result: if every $100 of rebate results in a sales increaseof less than 6.7% then the optimal policy is to offer no rebate. To see this, note that bythe formula derived in part (b), the optimal rebate is x = 750-1/(2e), which decreases tozero as e decreases to 1/(2*750) = 0.00067.2. In the pig problem, perform a sensitivity analysis based on the cost per day of keepingthe pig. Consider both the effect on the best time to sell and on the resulting profit. If anew feed costing 60 cents/day would let the pig grow at a rate of 7 lbs/day, would it beworth switching feed? What is the minimum improvement in growth rate that would makethis new feed worthwhile?In this case we havey = f (x) = (0.65 - 0.01 x) (200 + 7 x) - 0.60 xand so f '(x) = (195 - 14 x) / 100 = 0 at around x = 13.9, f (x) = 143.58. This is the globalmaximum since f (x) is a parabola. Since this is considerably more than the old maximumof 133.20, it is worth while to switch to the new feed.More generally we havey = f (x) = (0.65 - 0.01 x) (200 + g x) - 0.60 xwhere currently g = 7. Compute that f '(x) = (65 (g - 4) - 2 g x) / 100 = 0 whenx = 65 (g - 4) / (2 g)

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and substitute this back into the expression for y to obtainy = 13 (13 g^2 + 56 g + 208) / (16 g)which is equal to 133.20 when g = 5.26. Thus the new feed would be worth 60 cents perday as long as it produced a growth rate of at least 5.26 pounds per day.3. Reconsider the pig problem, of Example 1.1, but now assume that the price for pigs isstarting to level off. Letp = 0.65 - 0.01t + 0.00004 t^2(4)represent the price for pigs (cents/lb) after t days.(a) Graph Eq. (4) along with our original price equation. Explain why our original priceequation could be considered as an approximation to Eq. (4) for values of t near zero.The following graph shows the new, nonlinear price function along with the original, linearprice function. The original equation is the tangent line to the new equation at t = 0, andso the original equation may be considered as an approximation of the new equation forvalues of t near zero.GRAPHING UTILITYp = 0.65 - 0.01t + 0.00004 t^2-0.4-0.3-0.2-0.10.00.10.20.30.40.50.60.70102030405060708090100ptnew p(t)original p(t)(b) Find the best time to sell the pig. Use the five-step method, and model as a one-variable optimization problem.Step 1: Ask the question.Exactly the same as in the text p. 5, but now we assume p = 0.65 - 0.01 t + 0.00004 t^2.Step 2: Select the modeling approach.

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We will model this problem as a one variable optimization problem. See text p. 6.Step 3: Formulate the model.Let x=t and y=P, and writey = f(x) = (0.65 - 0.01 x + 0.00004 x^2) (200 + 5 x) - 0.45 x.Our goal is to maximize f(x) over the set of all x >= 0.Step 4: Solve the model.Compute that f '(x) = (3 x^2 - 420 x + 4000)/5000 which equals zero at x = 70-10*SQRT(321)/3 and at x = 70+10*SQRT(321)/3, or at around x = 10.28 and x =129.72. As shown in the following graph, the maximum is at x = 10.28. Although f(x)increases to infinity as x gets large, this is beyond the range where our model makes sense.GRAPHING UTILITYy = f(x)-40-20020406080100120140020406080100120140160yxStep 5: Answer the question.According to this model, the optimal policy is to sell the pig after 10 days, for a net profitof about $134.(c) The parameter 0.00004 represents the rate at which price is leveling off. Conduct asensitivity analysis on this parameter. Consider both the optimal time to sell and theresulting profit.Generalize the model from part (b) and let

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y = f(x) = (0.65 - 0.01 x + a x^2) (200 + 5 x) - 0.45 xwhere currently a = 0.00004. Compute thatf '(x) = (150*a*x^2+x*(4000*a-1)+8)/10 = 0atx = -(SQRT(16000000*a^2-12800*a+1)+4000*a-1)/(300*a).Thendx/de = -(SQRT(16000000*a^2-12800*a+1)+ 6400*a-1)/(300*a^2*SQRT(16000000*a^2-12800*a+1))and so S(x,e) = (dx/de) (e/x) = 0.31. If price is leveling off 10% faster then we thought,then we should wait 3.1% longer to sell the pig . A similar computation yields S(y,e) =0.008 so that if price is leveling off 1000% faster then we thought, then we should wait8% longer to sell the pig.(d) Compare the results of part (b) to the optimal solution contained in the text. Commenton the robustness of our assumptions about price.There is not much difference between the results in part (b) and the results in the text.Our model is reasonably robust with respect to the assumption that price is a linearfunction of time. Given this, the added computational difficulty associated with thequadratic price model is probably not justified.4. An oil spill has fouled 200 miles of Pacific shoreline. The oil company responsiblehas been given 14 days to clean up the shoreline, after which a fine will be levied in theamount of $10,000/day. The local clean-up crew can scrub 5 miles of beach per week ata cost of $500/day. Additional crews can be brought in at a cost of $18,000 plus$800/day for each crew.(a) How many additional crews should be brought in to minimize the total cost to thecompany? Use the five-step method. How much will the clean-up cost?Step 1: Ask the question.Variables:c= number of additional crewst= time to clean up spill (days)T= total cost of clean-up ($)F= fine ($)Assumptions:T= 500 t + (18000 + 800 t) c + F200= (5 / 7) (c + 1) tF= 0if t <= 14F= 10,000 (t - 14)if t > 14c is a nonnegative integer, and t >= 0Objective:Minimize T.Step 2: Select the modeling approach.

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We will model this problem as a one variable optimization problem. See text p. 6.Step 3: Formulate the model.Let x = c and y = T, and writey = f(x) = 500 (280 / (x+1)) +(18,000 + 800 (280 / (x+1))) xif x >= 19 ory = f(x) = 500 (280 / (x+1)) +(18,000 + 800 (280 / (x+1))) x+ 10,000 (280 / (x + 1) - 14)if x < 19. Our goal is to maximize f(x) over the set of nonnegative integers x.Step 4: Solve the model.One way to solve is to minimize over the nonnegative reals, and then specialize to theintegers. On 0 <= x < 19 we havef '(x) = 2000*(9*x^2+18*x-1349)/(x+1)^2which is negative on [0, 11.28) and positive on (11.28, 19), so x = 11.28 is the minimum.Then the minimum over the integers occurs at either x = 11 or at x = 12, and we cancheck that x = 11, f (x) = 508333 is smaller. On x >= 19 we havef '(x) = 6000*(3*x^2+6*x+17)/(x+1)^2which is always positive, and so on this interval the minumum occurs at x = 19, f (x) =561800. Then the global minimum occurs at x = 11.GRAPHING UTILITYy = f(x)05000001000000150000020000002500000300000005101520253035404550yxStep 5: Answer the question.

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According to this model, the optimal policy is to bring in 11 additional crews, resulting ina total clean-up cost of around $510,000. Clean-up will take around 23.3 days and theresulting fine will be around $93,000.(b) Examine the sensitivity to the rate at which a crew can clean up the shoreline.Consider both the optimal number of crews and the total cost to the company.Generalize the model in part (a) to obtain the assumption200= r (c + 1) twhere currently r = (5 / 7) miles per day per crew. Then we havey = f(x) = 500 (200 / (r (x+1))) +(18,000 + 800 * 200 / (r (x+1))) xif x >= 19 ory = f(x) = 500 (200 / (r (x+1))) +(18,000 + 800 * 200 / (r (x+1))) x+ 10,000 (200 / (r (x+1)) - 14)if x < 19. For values of r near (5 / 7) the minimum should still occur on the interval(0, 19). Solving0 = f '(x) = 2000*(9*r*x^2+18*r*x+9*r-970)/(r*(x+1)^2)yieldsx = (SQRT(970)-3*SQRT(r))/(3*SQRT(r))and sodx / dr = -SQRT(970)/(6*r^(3/2)).Substituting r = (5 / 7) we obtainS(x, r) = (dx / dr) (r / x) = -0.54so that if the cleanup crews are 10% faster than expected, the optimal number of crewsdecreases by about 5.4%. If we substitute the formula for the optimal x in terms of r intothe formula for y = f(x) in the case x < 19, we obtainy = -2000*(79*r-6*SQRT(970)*SQRT(r)-80)/rand then we can also calculatedy / dr = -2000*SQRT(10)*(3*SQRT(97)*SQRT(r)+8*SQRT(10))/r^2Substituting r = (5 / 7) we obtainS(y, r) = (dy / dr) (r / y) = -0.88so that if the cleanup crews are 10% faster than expected, the total cost of clean-updecreases by about 8.8%.(c) Examine the sensitivity to the amount of the fine. Consider the number of days thecompany will take to clean up the spill and the total cost to the company.Generalize the model in part (a) to obtain the assumptionF= a (t - 14)if t > 14where currently a = 10,000 dollars per day. Then for values of a near 10,000 we havey = f(x) = 500 (280 / (x+1)) +(18,000 + 800 (280 / (x+1))) x+ a (280 / (x + 1) - 14)if x < 19.The optimal number of crews is x = SQRT(14)*SQRT(a-300)/30-1 and so thetime to finish the clean-up is

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t = 280 / (x+1) = 600*SQRT(14)/SQRT(a-300)and then we can compute that at a=10,000 we haveS(t,a) = (dt/da)(a/t) = (-300*SQRT(14)/(a-300)^(3/2))*(a/t) = -0.52so that if the fine is raised by 2% then the cleanup time should decrease by about 1%.Substituting the optimal formula for x in terms of a into the equation for y above, weobtainy= 2*(600*SQRT(14)*SQRT(a-300)-7*a+103000)dy/da = 2*SQRT(7)*(300*SQRT(2)-SQRT(7)*SQRT(a-300))/SQRT(a-300)so that at a = 10,000 we haveS(y,a) = (dy/da)(a/y) = 0.17which means that if the fine is increased then the total cleanup cost to the company will goup by about 1.7% for each additional $1,000 per day of fine.(d) The company has filed an appeal on the grounds that the amount of the fine isexcessive. Assuming that the only purpose of the fine is to motivate the company to cleanup the oil spill in a timely manner, is the fine excessive?Reasonable answers will differ on this question. On the one hand, the fine is only 19% ofthe total cost, and if the fine were reduced by 50% then the number of days to clean up thespill would increase by about 25% and it would only save the company about 8.5% of thetotal cost. So the amount of the fine does not seem excessive. On the other hand, if the14 day limit were extended to 21 days, cleanup would proceed exactly as before, only thecompany would save $70,000. So in this case the 14 day limit does seem excessive.5. It is estimated that the growth rate of the fin whale population (per year) isr x (1 - x / K), where r = 0.08 is the intrinsic growth rate, K = 400,000 is the maximumsustainable population, and x is the current population, now around 70,000. It is furtherestimated that the number of whales harvested per year is about .00001 E x, where E isthe level of fishing effort in boat-days. Given a fixed level of effort, population willeventually stabilize at the level where growth rate equals harvest rate.(a) What level of effort will maximize the sustained harvest rate? Model as a one-variable optimization problem using the five-step method.Step 1: Ask the question.Variables:x= population (whales)E= level of effort (boat-days)g= growth rate (whales per year)h= harvest rate (whales per year)Assumptions:g= 0.08 x (1 - x / 400,000)h= .00001 E xg= h,x >= 0,E >= 0

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Objective:Maximize h.Step 2: Select the modeling approach.We will model this problem as a one variable optimization problem. See text p. 6.Step 3: Formulate the model.Let y = h, and writey = f(x) = 0.08 x (1 - x / 400,000).Our goal is to maximize f(x) over the interval x >= 0.Step 4: Solve the model.Compute f '(x) = 0 at x = 200,000, f(x) = 8000 and since the graph of f(x) is a parabolawe know this is the global maximum.Step 5: Answer the question.According to this model, the optimal policy is to harvest 8000 whales per year, whichrequires controlling the level of effort at 4000 boat-days per year. This will maintain thepopulation of whales at 200,000 which is higher than the current population. Thisindicates that in the past the harvesting rate has exceeded the optimum level according tothis model.(b) Examine the sensitivity to the intrinsic growth rate. Consider both the optimum levelof effort and the resulting population level.Generalize the model in part (a) to obtain the assumptiong= r x (1 - x / 400,000)where currently r = 0.08. Then we havey = f(x) = r x (1 - x / 400,000)and the optimum is still at x = 200,000 but now f(x) = 100,000 r which leads toE = 50,000 r. ThenS(x , r) = 0S(E , r) = 1because x does not depend on r, and E is proportional to r.(c) Examine the sensitivity to the maximum sustainable population. Consider both theoptimum level of effort and the resulting population level.Generalize the model in part (a) to obtain the assumptiong= 0.08 x (1 - x / K)where currently K = 400,000. Then we havey = f(x) = 0.08 x (1 - x / K)

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and the optimum is at x = K / 2, f(x) = 0.02 K which leads toE = 4000. ThenS(x , K) = 1S(E , K) = 0because E does not depend on K, and x is proportional to K.6. In problem 5, suppose that the cost of whaling is $500 per boat-day, and the price of afin whale carcass is $6,000.(a) Find the level of effort that will maximize profit over the long term. Model as a one-variable optimization problem using the five-step method.Step 1: Ask the question.Variables:x= population (whales)E= level of effort (boat-days)g= growth rate (whales per year)h= harvest rate (whales per year)R= revenue (dollars per year)C= cost (dollars per year)P= profit (dollars per year)Assumptions:g= 0.08 x (1 - x / 400,000)h= .00001 E xR= 6000 hC= 500 EP= R - Cg= h,x >= 0,E >= 0Objective:Maximize P.Step 2: Select the modeling approach.We will model this problem as a one variable optimization problem. See text p. 6.Step 3: Formulate the model.Setting g = h we see thatE = 0.08 (1 - x / 400,000) / .00001= 8000 - .02 xLet y = P, and writey = f(x) = 6000 (.00001 E x) - 500 E= (.06 x - 500) E= -.0012 x^2 + 490 x - 4,000,000.Our goal is to maximize f(x) over the interval x >= 0.

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Step 4: Solve the model.Compute f '(x) = 0 at x = 204,167, f(x) = 4.60208*10^7 and since the graph of f(x) is aparabola we know this is the global maximum.Step 5: Answer the question.According to this model, the optimal policy is to harvest 7997 whales per year, whichrequires controlling the level of effort at 3917 boat-days per year. This will maintain thepopulation of whales at 204,167 and will net the industry an annual profit of around 46million dollars.(b) Examine the sensitivity to the cost of whaling. Consider both the eventual profit in$ / year and the level of effort.Generalize the model in part (a) to obtain the assumptiony = f(x) = 6000 (.00001 E x) - w E= (.06 x - w) E= -.0012 x^2 + (480+.02 w) x - 8000 w.where currently w = 500. Then the optimum is at x = 200,000 + 25w/3,P = f(x) = (1/12) (w - 24000)^2 which leads to E = (24000 - w)/6. ThenS(P , w) = (dP/dw) (w/P) = -0.04S(E , w) = (dE/dw) (w/E)= -0.02so if the cost of whaling goes up by 10% then the optimal profit decreases by 0.4% andthe optimal level of effort decreases by 0.2 %.(c) Examine the sensitivity to the price of a fin whale carcass. Consider both profit andlevel of effort.Generalize the model in part (a) to obtain the assumptiony = f(x) = c (.00001 E x) - 500 E= (.00001 c x - 500) E= -(c/5,000,000) x^2 + (2c/25 + 10) x - 4,000,000.where currently c = 6000. Then the optimum is at x = 200,000 (c + 125)/c,f(x) = 8000 ( c^2 - 250 c + 15625) / c which leads to E = 4000(c-125)/c. ThenS(P , c) = (dy/dc) (c/y) = 1.04S(E , c) = (dE/dc) (c/E) = 0.02so if the price of a fin whale carcass goes up by 10% then the optimal profit increases byabout 10.4% and the optimal level of effort increases by 0.2 % .(d) Over the past 30 years there have been several unsuccessful attempts to ban whalingworldwide. Examine the economic incentives for whalers to continue harvesting. Inparticular, determine the conditions (values of the two parameters: cost per boat-day andprice per fin whale carcass) under which harvesting the fin whale produces a sustainedprofit over the long term.

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From part (b) we see that the optimal profit is P = f(x) = (1/12) (w - 24000)^2 at $6000per carcass, so that the industry makes a profit whenever the cost of whaling is below$24,000 per boat-day. From part (c) we see that P = 8000 ( c^2 - 250 c + 15625) / cat $500 per boat-day, in which case the industry makes a profit whenever the price percarcass exceeds $125. It is difficult analytically to consider the optimal P as a function ofboth c and w together, but using our sensitivity results we have approximately thatP = 46,000,000 + 1.04 (46,000/6) (c-6,000) - 0.04 (460,000/5) (w-500)and then P>0 whenever c > 6w/13 or in other words the industry makes a profit as long asthe price of a fin whale carcass is a bit more than half of the cost per boat-day of whaling.Thus there is a very strong profit motive to continue whaling.7. Reconsider the pig problem of Example 1.1, but now suppose that our objective is tomaximize our profit rate ($/day). Assume that we have already owned the pig for 90 daysand have invested $100 in this pig to date.(a) Find the best time to sell the pig. Use the five-step method, and model as a one-variable optimization problem.Step one is the same as in figure 1.1 of the text, except that we add a new variable Q =profit per day ($/day), we assume C = 100 + 0.45 t, Q = P / (t + 90), and our objective isto optimize profit per day. This is a one variable optimization problem. Letting x = t andy = f (x) = Q we are to find the maximum of the functionf (x) = ((200 + 5 x) (0.65 - 0.01 x) - (100+ 0.45 x)) / (x + 90)over the set of all nonnegative x.The graph indicates a maximum around x = 4.5,f (x) = 0.345.GRAPHING UTILITYy = f(x)0.3300.3320.3340.3360.3380.3400.3420.3440.346012345678910yxCompute

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f '(x) = -(x^2+180*x-840)/(20*(x+90)^2) = 0at x = 2*SQRT(2235)-90 or approximately x = 4.55. Then the farmer should sell the pigin 4 or 5 days to maximize profit per day, or the rate at which the farmer earns income.(b) Examine the sensitivity to the growth rate of the pig. Consider both the best time tosell and the resulting profit rate.Let g denote the growth rate of the pig, where currently we assume g = 5 lbs / day. Nowwe are to find the maximum of the functionf (x) = ((200 + g x) (0.65 - 0.01 x) - (100+ 0.45 x)) / (x + 90)over the set of all nonnegative x.Computef '(x) = -(g*x^2+180*g*x-150*(39*g-167))/(100*(x+90)^2) = 0atx = 5*SQRT(6)*(SQRT(93*g-167)-3*SQRT(6)*SQRT(g))/SQRT(g)and then at g = 5 we haveS(x, g) = (dx / dg) (g / x) = 4.82so that if the pig grows 1% faster than expected, we should wait 5% longer to sell the pig.Substituting into y = f(x) we can also compute thatS(y, g) = (dy / dg) (g / y) = 0.42so that if the pig grows 10% faster then expected we should gain an additional 4% profitper day.(c) Examine the sensitivity to the rate at which the price for pigs is dropping. Considerboth the best time to sell and the resulting profit rate.Let r denote the rate at which price is falling, where currently r = 0.01 ($/day). Then weare to maximizef (x) = ((200 + 5 x) (0.65 - r x) - (100+ 0.45 x)) / (x + 90)over the set of all nonnegative x.Computef '(x) = -(5*r*x^2+900*r*x+6*(3000*r-37))/(x+90)^2 = 0atx = SQRT(30)*(SQRT(3750*r+37)-15*SQRT(30)*SQRT(r))/(5*SQRT(r))and then at r = 0.01 we haveS(x, r) = (dx / dr) (r / x) = -5.16so that if the price drops 1% faster than expected, we should sell the pig 5% sooner.Substituting into y = f(x) we can also compute thatS(y, r) = (dy / dr) (r / y) = -0.31so that if the price drops 10% faster then expected then we will lose about 3% of ourexpected profit per day.8. Reconsider the pig problem of Example 1.1, but now take into account the fact that thegrowth rate of the pig decreases as the pig gets older. Assume that the pig will be fullygrown in another five months.

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(a) Find the best time to sell the pig in order to maximize profit. Use the five-stepmethod, and model as a one-variable optimization problem.The results of step one are the same as in figure 1.1 of the text, except that now weassume the growth rate of the pig is r (lbs/day) where r = 5 - t / 30 so that the weightw (lbs) of the pig after t days is w = 200 + (5 - t / 30) t. Then we need to maximizef (x) = (200 + (5 - x / 30) x) (0.65 - 0.01 x) - 0.45 xover the set of all nonnegative x.The graph indicates a maximum around x = 6,f (x) = 132.GRAPHING UTILITYy = f (x)130.0130.5131.0131.5132.0132.5012345678910yxWe compute thatf '(x) = (3*x^2-430*x+2400)/3000 = 0at x = 215/3-5*SQRT(1561)/3 = 5.82 so that f (x) = 132.29. Then the farmer should sellthe pig after 6 days, and the expected net profit will be about $132.(b) Examine the sensitivity to the time it will take until the pig is fully grown. Considerboth the best time to sell and the resulting profit.We generalize our previous model. Let a denote the rate at which the growth of the pigslows. Currently a = 1.0 lbs/day per month. Now the problem is to maximizef (x) = (200 + (5 - a x / 30) x) (0.65 - 0.01 x) - 0.45 xover the set of all nonnegative x.For values of a near 1.0 the maximum should occur at apoint near x = 6 where f '(x) = 0. We compute thatf '(x) = (3*a*x^2-10*x*(13*a+30)+2400)/3000 = 0at x = -5*(SQRT(169*a^2+492*a+900)-13*a-30)/(3*a). Then at a = 1.0 we haveS(x, a) = (dx / da) (a / x) = -0.28

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Solution Manual for Mathematical Modeling, 4th Edition - Page 16 preview image

so that if the pig stops growing 10% sooner than expected, then we should sell the pig 3%sooner. Substituting into the formula for y = f (x) we may also compute thatS(y, a) = (dy / da) (a / y) = -0.005so that the resulting profit is almost totally insensitive to the rate at which the pig stopsgrowing. This makes sense because the growth rate of the pig will not change much in the6 or so days until we sell.9. A local daily newspaper with a circulation of 80,000 subscribers is thinking of raisingits subscription price. Currently the price is $1.50 per week, and it is estimated that thepaper would lose 5,000 subscribers if the rate were to be raised by 10 cents/week.(a) Find the subscription price that maximizes profit. Use the five-step method, andmodel as a one-variable optimization problem.Step 1: Ask the question.Variables:p= subscription price ($/paper)s= number of subscriptions (papers)P= profit ($)Assumptions:s= 80000 - 50000 (p - 1.50)P= p ss>= 0p>= 0Objective:Maximize P.Step 2: Select the modeling approach.We will model this problem as a one variable optimization problem. See text p. 6.Step 3: Formulate the model.Let x = p and y = P, and writey = f(x) = x (80000 - 50000 (x - 1.50)).Our goal is to maximize f(x) over the interval [0, 3.10] since these are the only values of xthat satisfy both s >= 0 and p >= 0. In other words, price cannot be negative, andaccording to our model the number of subscriptions drops to zero when price is raised to$3.10.Step 4: Solve the model.A graph of the function f (x) shows that the maximum occurs at around x = 1.5 andf (x) = 120,000.

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