MAC 2233/-6 Business Calculus, Spring 2 Final Eam Name: Date: 5/3/2 Time: :am-2:nn Section: Show ALL steps. One hundred points equal % Question. (8 points) The following diagram shows the graphs of eight equations. Match each of the following equations with its graph. (Eplanation is not needed.) (a) y = +. Answer: B (b) y = 2 +4 +. Answer: H (c) y = 3e/5 +e /5. Answer: E (d) y = 9 2. Answer: A (e) y = 6 2/3 8/3. Answer: F (f) y = 4 2 + 6 2. Answer: D (g) y = ( 4)( )( + 3). Answer: C (h) y = ln() + ln(3/). Answer: G Question 2. (3+2 points) Mrs. Rose starts a business selling pâté to friends and neighbors. She spends $6 to buy equipment (food processor, pots and pans). She charges $7 per serving. Her costs include $ fied cost per month plus $3.5 per serving. She estimates that she can sell 4 servings each month in the first year of her business. (a) Find the monthly profit P of her pâté business. Solution. The monthly revenue is R = 4 7 = $28. Her monthly cost is C = + 3.5 4 = $24. Her monthly profit is P = R C = 28 24 = $4. (b) How many months does it take for Mrs. Rose to break even? Solution. In order to break even, her profits after months should equal her initial cost of setting up the business (which is $6 in this case). Thus, 4 = 6, giving that = 5 months. Keeping Scores: Question 2 3 4 5 6 7 8 9 2 3 Total Score Out of 8 5 4 9 2 6 6 6 8 8 23
Question 3. (2+4+4+4 points) Find the limits below analytically. If the limit does not eist, check if it is + or. Show all work. (a) lim ( 3 2 + 3 + ). Solution. Direct substitution gives 3 2+ 3 + = 2+2 =. Thus, direct substitution works, and the answer is lim (3 2 + 3 + ) =. (b) lim 3 2 9. Solution. Direct substitution (of = 3) gives 3 is + or, we see that 2 9. Thus the limit does not eist. To check if the answer < when < 3. Thus, lim 3 2 9 =. + (c) lim. Solution. Direct substitution gives /, and we need to rationalize the numerator: + = + + + + + = ( + ) ( + + ) =. + + Thus, lim + = lim = + + 2. Note that we have obtained the derivative of f() = directly from the limit definition of derivatives. 4 (d) lim 2 + +. 3 Solution. As, we have 4 2 + + 3 4 2 3 = 2 3 = 2 3 = 2 3. Thus, lim 4 2 + + 3 = 2 3. 2
Question 4. (3+4+4+4+4 points) Find the derivatives of the following functions. There is no need to simplify your answers. (a) f() = + 3/2 5 + e + ln(2). Solution. We rewrite the function as f() = /2 + 3/2 5+e +ln(2)+ln() and then differentiate to get f () = 2 3/2 + 3 2 /2 + e +. (b) f() = 2e 2 3. Solution. We use the product rule to get (c) f() = ln( ). Solution. We use the quotient rule to get ( (d) y = ln e e + ). f () = (2)e 2 3 + (2)e 2 3 (2) = 2(2 2 + )e 2 3. f () = Solution. We first rewrite as ln( ) 2 = y = 2 (ln(e ) ln(e + )), ( )ln( ) 2. ( ) and then differentiate to obtain y = ( ) e 2 e e e = e + e 2. (e) y = (ln() + 3) 2. Solution. This a direct application of the chain rule: y = 2 (ln() + 3) = 2 (ln() + 3). 3
Question 5. (3+4+4 points) Find the indefinite integrals. Show all steps. (a) ( 2 3 + 4 4 3 + e ) d. Solution. We integrate term-by-term to obtain ( 2 3 + 4 4 3 + e ) d = ln + 2 + 4 4 + e + C. (b) d. (ln()) 2 Solution. We let u = ln(). Then du d =, and we use the general power rule for integrals: (ln()) 2 d = (c) e 3 e 3 + d. Solution. Let u = e 3 +. Then du d = 3e3. Thus, e 3 e 3 + d = 3 (ln()) 2 d = (ln()) + C = ln() + C 3e 3 e 3 + d = 3 ln(e3 + ) + C. Question 6. (4+4+4 points) Evaluate the definite integrals. Show all steps. (a) ( )( + ) d. Solution. We epand the integrand, and then integrate (b) 2 2 + ln(2 + ) d. ( 2 ) d = [ 3 3 ] = = 2 3. Solution. Let u = ln( 2 + ). Then du d = 2 2 +. Hence, [ ] 2 2 + ln(2 + ) d = 2 (ln(2 + )) 2 = 2 (ln(2))2.24. (c) Find the area of the region enclosed by y = 2 +, =, = 2 and the -ais. Solution. The area equals the definite integral: 2 ( 2 + ) d = [ 3 3 + 2 2] 2 = 3 8 + 2 = 4 3. 4
Question 7. (6 points) Find the slope of the tangent line to the graph of the equation e y + 2 = y 2 + 2 at the point (, ). Solution. We use implicit differentiation to obtain e y ( y + y ) + 2 = 2yy. We then substitute = and y = to get ( + y ) + 2 = = y = 2. Question 8. (3+3 points) (a) Find the eponential function y = Ce kt (or y = C b t/t ) that passes through the given two points: (, 5) and (4, 3). Solution. Since y = 3 when t =, we have C = 5. We net substitute t = 4 and y = 3 to get 3 = 5e 4k = k = 4 ln(.6) =.277. Thus, y = 5e.277t. Or we can proceed as follows. Since y = 5 when t =, we have C = 5. In addition, y is multiplied by 3 5 after 4 units of times. Thus, y = 5 (3 5) t/4. (b) For the eponential model found in (a), how long does it take for the value of y to be halved? Solution. We solve for t when y = 5 2 : 5 2 = 5e.277t = t =.277 ln(.5) = 5.428 units of time. Or if we use the other model, we solve for t: 5 2 = 5 (3) t/4 5 = t = 4 ln(.5) ln(.6) = 5.428 units of time. Question 9. (2+4 points) The marginal cost (in dollars) for selling packets of rechargeable batteries and their charger is dc d = 8.. (a) Use the given information to estimate the change in C if current production is increased from to 5. Solution. Note that the marginal cost gives an approimate change in cost for each additional packet sold. Thus, the change in profit is approimated by C dc d = (8. ) 5 = $35. (b) Find the eact change in cost when production is increased from to 5. Solution. For the eact change in cost, we need to calculate C = C(5) C(). We use integration: C = 5 dc d d = 5 (8.) d = [ 8.5 2] 5 = 784.875 75. = $34.875. 5
Question. (2+++2+2+2 points) Mrs. Rose owns a business selling pâté locally. She would like to evaluate her business through the following eercise. (a) She estimates that she will sell = 5 servings in a month for a price of p = $8 per serving, and she will sell = 4 servings when p = $9. Assume that the demand for her pâte is linear. Find the demand equation relating p to. Solution. From her estimates, we have Using the point-slope formula, we have p = 8 9 5 4 =.. p 8 =.( 5) = p =. + 3. (b) Find her monthly revenue R as a function of. Solution. R = p = (. + 3) =. 2 + 3. (c) Her monthly cost is C = + 4. Find her monthly profit P as a function of. Solution. P = R C = (. 2 + 3) ( + 4) =. 2 + 9. (d) Mrs. Rose is currently selling = 4 servings a month. Evaluate dp d at = 4. Solution. The marginal profit is dp =.2 + 9. d Thus, at = 4, the value of the marginal profit is dp d =.2 4 + = $ per serving. (e) What is the meaning of the numerical value found in (d)? Solution. The marginal profit is one dollar per serving. It means that her monthly profit will increase by about one dollar for each additional serving that she sells. (f) For the purpose of increasing her profit, Mrs. Rose is gradually lowering the price of her pâté. She is currently selling = 4 servings, and her sales is increasing at a rate of servings per month. Find the rate of change of her profit. Solution. From (d), we have dp d =. It is given that d dt dp dt = dp d = = $ per month. d dt =. Her rate of change of profit is therefore 6
Question. (4+4 points) Mr. Jones owns a small local engineering consulting business. The annual profit of his business from 2 to 27 is shown in the chart, where P is the annual profit in thousands of dollars, and t = means year 2. (a) Estimate the average rate of change of P from t = to t = 7. Solution. The average rate of change equals the slope of the secant line between t = and t = 7. From the graph, we see that P = 2 when t =, and P = 47 when t = 7. Thus, the average rate of change between t = and t = 7 equals P t = 45 2 7 = 5 thousand dollars per year. (b) Estimate the rate of change of P at t = 5. Solution. The rate of change of P at t = 5 equals the slope of the tangent line at t = 5. Thus, we draw the tangent line at t = 5, and find two convenient points on the tangent line: (, 4) and (5, 22). The rate of change of P at t = 5 therefore equals P t = 22 4 = 4.5 thousand dollars per year. 5 Question 2. (4+4 points) Company XYZ wants to investigate the effect of advertisement on the revenue of a certain product. The following function is used to model the situation: R = 3 + 23 2 3 + 58, 5. where R is the monthly revenue in thousands of dollars, and is the amount of advertising in thousand dollars each month. (a) Find the value of where the point of diminishing return occurs. Solution. We differentiate R twice to obtain dr d = 3 2 + 23 5 3, d 2 R d 2 = 3 5 + 23 5. Interval (, 23 3 ) (23 3, 5) Test value Thus, the candidate (obtained by setting d2 R = ) is = 23 d 2 3. Sign of d2 R + d 2 We note that = 23 Concavity up down 3 is in the domain. To check that it indeed gives rise to a point of diminishing return, we use the table o the right. As there is a change of concavity at = 23 3, we see that there is a change of concavity at = 23 3. Thus, = 23 3 thousand dollars. (b) Eplain the meaning and significance of the point of diminishing return. dr Solution. The point of diminishing return is where the marginal d changes from increasing to d decreasing (that is, 2 R from positive to negative). Each investment dollar (on advertising in our d 2 case) beyond the point of diminishing return has a diminishing value than the previous dollar (since dr d is decreasing). Investment beyond the point of diminishing return is in general not considered a wise use of capital. 7
Question 3. (++2+6 points) A rancher plans to enclose a rectangular area for his livestock. The area is adjacent to a river and is shown in the diagram on the right. The rancher has $6 to spend on fencing the area. The sides net to the river do not need fencing, but the fencing on all other sides costs $ per yard. Let, y be the lengths (in yards) of the fences shown in the diagram, where the numbers associated with each side represent the costs per yard. By following the steps below, we aim to find the values of, y so that the area enclosed is maimum. (a) Find the cost equation relating, y to the amount of money ($6) available. Solution. Since 2 + 3y yards are required for fencing and the cost $ per yard, we have 2 + 3y = 6. y y y (b) Find an equation relating the enclosed area A with, y. Solution. As area equals length times width, where length is 2 and width is y in this problem, we have A = 2y. (c) Epress A as a function of one variable. Find also the domain of the function. Solution. From (a), we have 2 = 6 3y. Substituting into the equation in (b) to obtain A = (6 3y)y = 6y 3y 2. To find the domain, we note that y obviously and that y is maimum when =. Using (a), we see that y 2. Thus, the domain is [, 2]. (d) Use calculus to find, y so that A is maimum. Check for maimality of A. Solution. We first find that da da dy = 6 6y. For critical numbers, we set dy = and find that y = is the only critical number. Since A is a continuous function on a closed interval, we compare the values of A at the critical number as well as at the endpoints (of the interval). For this, we have the following table. y 2 A 3 Thus, A is maimum (with A = 3 square yards) when y = yards. To find the corresponding value of, we see that 2 = 6 3y = 3, or = 5. Thus, the solution is = 5 yards and y = yards. 8