BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.2, Problem 50E

(a)

To determine

To describe: The meaning of f(20)=10,000 and f(20)=350.

Expert Solution

Explanation of Solution

The value f(20)=10,000 means that if the selling price is 20 dollars per yard, then 10,000 yards will be sold.

The value f(20)=350 means that increasing selling price by $1 then sales will decrease by 350 yards.

(b)

To determine

To find: The value of R(20) and interpret.

Expert Solution

Explanation of Solution

Given:

The function R(p)=pf(p).

The values f(20)=10,000 and f(20)=350.

Derivative rule:

(1) Product Rule: ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

(2) Power rule: ddx(xn)=nxn1

Calculation:

Obtain the value of R(20).

The derivative of the function R(p)=pf(p) is computed as follows,

R(p)=ddp(pf(p))

Apply the product rule (1) and the power rule (2),

R(p)=pddp(f(p))+f(p)ddρ(p)=pf(p)+f(p)(1)=pf(p)+f(p)

Thus, the derivative of the function R(p)=pf(p) is R(p)=pf(p)+f(p).

Substitute 20 for p in R(p)=pf(p)+f(p),

R(20)=20f(20)+f(20)

Substitute the value f(20)=10,000 and f(20)=350 in R(20),

R(20)=20(350)+10,000=7000+10,000=3000

The value of R(20) is positive.

That is, if the value of selling price is increase then the revenue is increases.

Therefore, the Revenue increases instaneously by $3000 for every $1 increase in price per yard.

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