Chapter 3.2, Problem 50E

A manufacturer produces bolts of a fabric with a fixed width. The quantity q of this fabric (measured in yards) that is sold is a function of the selling price p (in dollars per yard), so we can write q = f(p). Then the total revenue earned with selling price p is R(p) = pf(p).

(a) What does it mean to say that f(20) = 10,000 and f'(20) = –350?

(b) Assuming the values in part (a), find R'(20) and interpret your answer.

To determine

To describe: The meaning of $f\left(20\right)=10,000$ and $f^{\prime }\left(20\right)=-350$.

Explanation of Solution

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

The value $f^{\prime }\left(20\right)=-350$ means that increasing selling price by $1 then sales will decrease by 350 yards.

To determine

To find: The value of $R^{\prime }\left(20\right)$ and interpret.

Explanation of Solution

Given:

The function $R\left(p\right)=pf\left(p\right)$.

The values $f\left(20\right)=10,000$ and $f^{\prime }\left(20\right)=-350$.

Derivative rule:

(1) Product Rule: $\frac{d}{dx}\left[f_1\left(x\right)f_2\left(x\right)\right]=f_1\left(x\right)\frac{d}{dx}\left[f_2\left(x\right)\right]+f_2\left(x\right)\frac{d}{dx}\left[f_1\left(x\right)\right]$

(2) Power rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Calculation:

Obtain the value of $R^{\prime }\left(20\right)$.

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

$R^{\prime }\left(p\right)=\frac{d}{dp}\left(pf\left(p\right)\right)$

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

$\begin{array}{c}{R}^{\prime }\left(p\right)=p\frac{d}{dp}\left(f\left(p\right)\right)+f\left(p\right)\frac{d}{d\rho }\left(p\right)\\ =p{f}^{\prime }\left(p\right)+f\left(p\right)\left(1\right)\\ =p{f}^{\prime }\left(p\right)+f\left(p\right)\end{array}$

Thus, the derivative of the function $R\left(p\right)=pf\left(p\right)$ is $R^{\prime }\left(p\right)=p{f}^{\prime }\left(p\right)+f\left(p\right)$.

Substitute 20 for p in $R^{\prime }\left(p\right)=p{f}^{\prime }\left(p\right)+f\left(p\right)$,

$R^{\prime }\left(20\right)=20f^{\prime }\left(20\right)+f\left(20\right)$

Substitute the value $f\left(20\right)=10,000$ and $f^{\prime }\left(20\right)=-350$ in $R^{\prime }\left(20\right)$,

$\begin{array}{c}{R}^{\prime }\left(20\right)=20\left(-350\right)+10,000\\ =-7000+10,000\\ =3000\end{array}$

The value of $R^{\prime }\left(20\right)$ 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.