Page 234 - How To Solve Word Problems In Calculus

Solutions to Supplementary Problems

1. In this problem x represents the company’s advertising investment
(thousands of dollars) and D represents the demand for the
company’s product (thousands of items).The rate at which demand
is changing with respect to advertising dollars is D (x).

2
D(x) = 2000x + 900x + 60
D (x) = 4000x + 900

D (1.5) = 4000(1.5) + 900 x = 1.5 corresponds to $1500

= 6900

Demand increases at the rate of 6900 items per $1000 spent on
advertising.

2. Let x = demand for gasoline x = 2000 − 100 p − .05 p 2
p = selling price

dp dx
Given: =−0.15 Find: when p = 1.25
dt dt
dx dx dp
= ·
dt dp dt

dp
= (−100 − 0.1p)
dt

dx
When p = 1.25, = (−100 − 0.125)(−0.15)
dt
= 15.01875

The family’s demand for gasoline increases at the rate of
15.02 gallons per month.

3. Let x = number of yachts produced
p = profit

dx dp
Given: = 4 Find: after 6 months
dt dt

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