Math 103. Homework 9. Solutions

Math 103. Homework 9. Solutions 1. [80, p. 317] As a result of increasing energy costs, the growth rate of the profit of the 4 year old Venice Glass-blowing Company has begun to decline. Venice's management, after consulting with energy experts, decides to implement certain energy conservation measures aimed at cutting energy bills. The general manager reports that, according to his calculations, the growth rate of Venice's profit should be on the increase again within 4 years. If Venice's profit (in hundreds of dollars) x years from now is given by the function P(x) = x3−9x2+40 x+50 (0 ≤ x ≤ 8) determine whether the general managers forecast will be accurate. Solution. The first derivative P′(x) = 3x2−18x+40 and the second P′′(x) = 6x−18=6(x−3) The sign graph for the second derivative is This shows that (3,116) is an inflection point. This analysis reveals that after declining for the first 3 years, the growth rate of the company's profit is again on the raise. 2.[35, p. 369] The weekly demand for video discs manufactured by the Herald Record Co. is given by p=−0.0005 x2 +60 where p denotes the unit price in dollars and x denotes the quantity demanded. The weekly total cost function associated with producing theses discs is given by C(x)=−0.001x2 +18 x +4000 where C(x) denotes the total cost incurred in pressing x discs. Find the production level that will yield a maximum profit for the manufacturer. What is this maximum profit? Solution. The revenue function is R(x)=xp=x(−0.0005x2 +60) = −0.0005x3 +60 x and the profit function P(x)=R(x)−C(x)=−0.0005x3 +60 x−(−0.001x2 +18 x +4000)=−0.0005x3 +0.001x2 +42 x −4000 The first derivative P′(x) = −0.0015 x2 +0.002 x +42 Setting P′(x)=0 and using the quadratic formula gives x=−166.667 and x=168 We reject the negative solution. Moreover, P′′(x) = 0.002 −0.003 x so that P(x) is concave down for x ≥ 2/3. This means that x=168 gives a maximum. Therefore, the required level of production is 168 video discs. 3. [52, p. 352] The quantity demanded per month of the Sicard wristwatch is related to the unit price by the equation p= 50 (0 ≤ x ≤ 20) 0.01 x2 +1 where p is measured in dollars and x in units of a thousand. How many watches must be sold to yield a maximum revenue? Solution. The revenue function is R(x) = xp = 50x 0.01 x2 +1 To find the maximum value of R(x) we compute the first derivative using the quotient rule R′(x) = (50)(0.01 x2+1)−50 x(0.02 x) (0.01 x2+1)2 = (0.5)(100−x2) (0.01 x2+1)2 Setting R′(x)=0 implies 100−x2=0 and thus x=±10. Only the positive root is meaningful, and so x=10 is the only critical point we need to consider. Next we compute x 0 10 20 R(x) 0 250 200 and conclude that R(10)=250 is the absolute maximum value of R(x). Thus the revenue is maximized by selling 10,000 watches. 4. [15, p. 366] If exactly 200 people sign up for a charter flight, the LWP Travel Agency charges $ 300 per person. However, if more than 200 people sign up for the flight (assume this to be the case) then each fare is reduced by $ 1 for each additional person. Determine how many passengers will result in a maximum revenue for the travel agency. What is the maximum revenue? What would be the fare per passenger in this case? Solution. Using the hint in the book, we want to maximize the function R(x) = (200+x)(300−x) = −x2 +100 x+60000 The derivative R′(x) = −2x+100 and setting R′(x)=0 it obtains x=50. Since the second derivative R′′(x)=−2, we see that R(x) is concave down, and deduce that x=50 gives the absolute maximum of R. Therefore the number of passengers should be 250. The fare will then be $250/ passenger and the revenue will be $62,500. 5. You wish to construct a closed rectangular box that has a volume of 4ft3. The length of the base of the box will be twice as long as its width. The material for the top and bottom of the box costs 30 cents/square foot. The material for the sides of the box costs 20 cents/square foot. Find the dimensions of the least expensive box that can be constructed. Solution. From the picture we see that the total cost is C(x) = 30(2)(2x)x) +20(2)(2x+h) = 120 x2 +120 xh The volume is V=x(2x) h=4 or h= 2 x2 Therefore ⎛2 ⎞ 240 2 C(x) = 120 x +120x ⎜ ⎟= 120x + x2 ⎝x2 ⎠ 2 The derivative is C′(x) = 240 x− 240 x2 Setting C′(x)=0 gives 240 x= 240 x2 or x3=1. Therefore x=1. Moreover, C′′(x)=240 + 480 x3 is positive for all x > 0, so that C(x) is concave up there. It follows that x=1 gives an absolute minimum. The dimensions which minimize cost are 1 ft x 2 ft x 2 ft.