MA119-A Applied Calculus for Business Fall Homework 10 Solutions Due 11/8/ :30AM

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1 MA9-A Applie Calculus for Business 006 Fall Homework 0 Solutions Due /8/006 0:30AM 53 #8 How long will it take $,000 to grow to $5,000 if the investment earns interest at the rate 8%/year compoune monthly? By the compoun interest formula, we have 5000 = :08 t We use log 0 to solve t The equation becomes + 0:08 t = = 5 4 By applying log 0 on both sies, we have 5 log 0 4 = log 0 + 0:08 Therefore, t = log t = t log 0 + 0:08 = :7986 log 0 + 0:08 So, it takes about years an 0:7986 = 9:583 0 months 53 #8 Saving Accounts Bernie investe a sum of money 5 yr ago in a savings account, which has since pai interest at the rate of 8%/yr compoune quarterly His investment is now worth $,89 How much i he originally invest? From the problem, we know that the eposit in the beginning P is unknown The annual rate r = 8% = 0:08 Since it is compoune quarterly, m = 4 Since we are talking about 5 years, we have t = 5 Also, the total amount is A = 89: Thus, we have 89: = A = P + r mt = P + 0:08 45 = P (:0) 0 m 4 Solve for P = 89: 0 = 5000 (:0) So, he originally invest $5,000 ollars

2 53 #40 E ect of In ation on Salaries Omar s current annual salary is $35,000 How much will he nee to earn 0 yr from now in orer to retain his present purchasing power if the rate of in ation over that perio is 6%/year? Assume that in ation is continuously compoune From the problem, we know that the amount in the beginning P = The annual rate r = 6% = 0:06 Since we are talking about 5 years, we have t = 0 Since it is compoune continuously, we have A = P e rt = 35000e 0: So, he nees to earn $63,774 ollars in orer to retain his present purchasing power 54 #4 Fin the erivative of the function The erivative is f (x) = e x x 54 # Fin the erivative of the function The erivative is f (x) = e x x ( x) = e x ( ) = e x f (x) = ex + e x x f (x) = x ex + e x x 54 #6 Fin the erivative of the function The erivative is s f (s) = f (s) = s + e s = ex e x s s + e s + s + e s s = (s) e s + s + e s s s = se s + s + e s ( s) = s 3 e s

3 54 #30 Fin the secon erivative of the function f (t) = 3e t 5e t The rst erivative is t f (t) = 3e t ( ) 5e t ( ) = 6e t + 5e t The secon erivative is t f (t) = 6e t ( ) + 5e t ( ) = e t 5e t 54 #44 Fin the absolute extrema of the function h (x) = e x 4 on [ ; ] First, we n where h 0 (x) = 0 We have h 0 (x) = e x 4 (x) = xe x 4 Note that e x 4 > 0 for all x If h 0 (x) = 0, then we have x = 0 Therefore, we have only one critical point x = 0 To get the absolute maximum an minimum, critical points an bounary points are all the caniates Thus, we have three caniates x = ; 0; now We have Caniates 0 Value of h (x) e 4 Maximum or Minimum Abs Max Abs Min Abs Max The graph looks like 3 y x 54 #66 Maximum Oil Prouction It has been estimate that the total prouction of oil from a certain oil well is given by T (t) = 000 (t + 0) e 0:t + 0; 000

4 4 thousan barrels t yr after prouction has begun Determine the year when the oil well will be proucing at maximum capacity First, we n where T 0 (t) = 0 We have T 0 (t) = 000 () e 0:t + (t + 0) e 0:t ( 0:) = 00te 0:t Note that T 0 means that the rate of change of the total prouction after t yr When T 0 is increasing, the oil well can still prouce more Thus, it is not proucing at maximum capacity yet When T 0 is ecreasing, the oil well can only prouce less the the previous moment It means that it passes the maximum capacity alreay So, we are looking for a time t that T 0 changes from increasing to ecreasing Thus, we nee to n where T 00 (t) = 0 We have T 00 (t) = 00 () e 0:t + te 0:t ( 0:) = 00 e 0:t + 0:te 0:t = 00e 0:t ( 0:t) Note that e 0:t > 0 for all t If T 00 (t) = 0, then we have 0:t = 0, or, t = 0 Therefore, we have only one critical point of T 00 at t = 0 Therefore, the 0th year is the turning point of the oil well from proucing more every year to proucing less every year Hence, 0th year is the year when the oil well is proucing at maximum capacity 55 #6 Fin the erivative of the function h (t) = ln t 5 The erivative is t h (t) = ln t5 = t t 5 55 #4 Fin the erivative of the function f (x) = ln x + x Note that The erivative is x f (x) = x ln x + x 55 #0 Fin the erivative of the function Note that = = ln (x + ) ln (x ) 5t 4 = 0 t (ln (x + ) ln (x )) = ln (x + ) x x + () x () = x + x f (x) = ln x ln x = 4 ln x 3 3 ln (x ) x

5 5 The erivative is x f (x) = x = 4 55 #6 Fin the erivative of the function Note that x 3 4 ln x3 3 = 4 ln x3 3 x 3 3x = x x 3 3 f (x) = ln p x 4 ln p x 4 = ln x 4 = ln x 4 The erivative is x f (x) = x ln x 4 = ln x 4 x = x 4 (x) = x x 4 55 #38 Use logarithmic i erentiation to n the erivative of the function y = (3x + ) 4 (5x ) By applying ln on both sies, we have ln y = ln (3x + ) 4 (5x ) = ln (3x + ) 4 + ln (5x ) = 4 ln (3x + ) + ln (5x ) By i erentiating both sies, we have y y0 = 4 3x + (3) + 5x Thus, the erivative is y 0 = y 3x x (5) = 3x x = (3x + ) 4 (5x ) 3x x 55 #4 Use logarithmic i erentiation to n the erivative of the function p 4 + 3x y = 3p x + By applying ln on both sies, we have p 4 + 3x ln y = ln 3p x + = ln p 4 + 3x ln 3p x + = ln 4 + 3x ln x + 3 = ln 4 + 3x 3 ln x +

6 6 By i erentiating both sies, we have y y0 = 4 + 3x (6x) 3 x + (x) = Thus, the erivative is p 3x y x = y = 4 + 3x 3x + 3 3p x + 3x 4 + 3x 3x + 3! 3x 4 + 3x 3x #58 Fin the extrema of the function g (x) = x on [; 5] ln x First, we n where g 0 (x) = 0 We have g 0 (x) = ()(ln x) (x)( x) = ln x Note that on (ln x) (ln x) the interval [; 5], ln x > 0 If g 0 (x) = 0, then we have ln x = 0 The root of this equation is x = e :788 Therefore, we have only one critical point x = e To get the absolute maximum an minimum, critical points an bounary points are all the caniates Thus, we have three caniates x = ; e; 5 now We have Caniates e 5 Value of g (x) :8854 e : :067 ln ln 5 Maximum or Minimum Abs Min Abs Max The graph looks like y x