MATH 180 FALL 1995

ASSIGNMENT #11

2.4, #8. If f′(10) = −29000, then a small change h in price from $10 to $(10+h) produces a corresponding change in the quantity sold from 240000 to 240000 − 29000h.

2.4, #11. The units of g′(55) are (miles/gallon)/(miles/hour). To say g′(55) = −0.54 is to say that at velocity v = 55 miles/hour, a small change h in velocity produces a corresponding change of −.54h miles/gallon in efficiency. So driving 57 miles/hour decreases efficiency by 1.08 mile/gallon; driving 53 miles/hour increases efficiency by 1.08 mile/gallon.

2.4, #14. (a) One would expect f′ to be positive, i.e., an increase in advertising costs produces an increase in revenues from sales. If f′(100) = 2, then a small change h in advertising costs when the costs are $100000 produces 2h in revenue. In other words, each $1000 added to $100000 produces $2000 more in revenue, each $1000 subtracted from $100000 produces $2000 less in revenue. On the other hand, if f′(100) = .5, then each additional $1000 added to $100000 produces $500 more in revenue, each $1000 subtracted from $100000 produces $500 less in revenue. Hence the company should spend slightly more than $100000 if f′(100) = 2 and slightly less if f′(100) = .5.

Comments: This assignment had lower scores than any of the previous assignments. A lot of misunderstandings about the meaning of f′(a) showed up. Here are samples of misinterpretations of “g′(55) = −0.54” in problem #11 taken from homework.

• “at 55 miles/hour, it takes just over 1/2 hour to use 1 gallon of gas”

• “When the car is travelling at a rate of 55 mph it is losing gas at a rate of −.54 gallons per mile”

• “That the efficiency decreases every time of .54 of hr^{2}”

• “while driving at the speed of 55 mph, the fuel efficiency gets less”

• “−0.54 is the rate of consumption of gasoline”

• “the car is using −.54 less gallons at 55 mph” Compare these with the answer given above.