Math 103. Homework 3. Solutions

1. The Ace Truck Leasing Co. leases a certain size truck at $30/day and $.15/mile, whereas the Acme Truck leasing Co. leases the same size truck at $25/day and $.20/mile.

(a) Find the daily cost of leasing from each company as a function of the number of miles driven and sketch the graph of these functions.
(b) Which company should you rent a truck from for 1 day if you plan to drive 70 miles and wish to minimize cost?

Solution. (a) The daily cost function for the Ace Truck leasing company is f(x) = (.15)x +30, where x is the number of miles driven. For the Acme Truck Co. the daily cost is g(x) = (.20)x +25.

(b) If you rent a truck for one day from the Ace Truck Company and drive 70 miles then you pay f(70) = (.15) 70 +30 = 40.5 dollars. If you rent from Acme Truck Co. you pay g(70) = (.20) 70 +25 = 39 dollars. If you wish to minimize the cost, then you should rent from Acme Truck Co.

2. The supply function for certain car model is given by p=(0.1)x²+(0.5)x+15, where x is the quantity supplied (in thousands) and p is the unit price (in thousand dollars). What unit price will induce the supplier to make 5,000 cars available in the market place?

If the demand function for this particular car model is given by p=(0.7)x+(14.9), what are the equilibrium quantity and equilibrium price?

Solution. (a) To find the unit price that will induce the supplier to make 5,000 cars available, plug in x=5 into p=(0.1)x²+(0.5)x+15 and obtain p=20, hence the supplier will be induced to make 5,000 cars available if the unit price is $20,000.

(b) Solve the equation supply = demand, which in this case is

(0.1)x2+(0.5)x+15=(0.7)x+(14.9).

It reduces to the equation (.1)x²-(.2)x +.1=0. It has one positive solution (by the quadratic formula) which is x=1. The corresponding value for p=15.6. The equilibrium quantity is 1,000 cars and the equilibrium price is $ 15,600.

3. Sketch the graph of the following function f(x) and evaluate the limit lim_{x® 1} f(x), if it exists.

f(x)= ì ï í ï î x if x < 1, 0 if x=1 2-x if x > 1

Solution. As x® 1 from the left, the values f(x)=x approach 1. As x® 1 from the right, the values f(x)=2-x also approach 1. Therefore the limit exists and

lim x® 1  f(x)=1.

4. Find the limit

lim x® 1  x3+1 2x3+2

Solution. The limit of the numerator is lim_{x® 1} x³+1 = 2 and that of the denominator is lim_{x® 1} 2x³+2=4. Since the limit of the denominator is ¹ 0, the properties of limit say that

lim x® 1  x3+1 2x3+2 = lim x® 1  x3+1 lim x® 1  2x3 +2 = 2 4 = 1 2

5. Find the limit, if it exists,

lim x® 1  x x-1

Solution. Plugging x=1 into the expression x/(x-1) gives 1/0, which means that the limit does not exist.