Chapter 1.7, Problem 46E

Correcting a Prediction The financial analysts at the store in Example 5 corrected their projections and are now expecting the total sales for the $x$

day of January to be

$T(x)=\frac{24}{5}+\frac{36}{5{(3x+1)}^2}$

thousand dollars.

a. Let $S(x)$

be as in Example 5. Compute $T(1),T\text{'}(1),S(1),$

and $S\text{'}(1)$.

b. Compare and interpret the data in part (a) as they pertain to the sales on January $1$.

Example 5:

Declining Sales:

At the end of holiday season in January, the sales at a department store are expected to fall (Fig. 3). It is estimated that for the $x$

day of January the sales will be

$S(x)=3+\frac{9}{{(x+1)}^2}$

thousand dollars.

a. Compute $S(2)$ and $S\text{'}(2)$ and interpret your results.

b. Estimate the value of sale on January $3$

and compare your result with the exact value $S(3)$

Solution:

a. We have, $S(2)=3+\frac{9}{{(2+1)}^2}=4$

thousand dollars.

To compute $S\text{'}(2)$, write $\frac{9}{{(x+1)}^2}$

as $9\cdot {(x+1)}^{-2}$, then,

$S\text{'}(x)=\frac{d}{dx}\left[S(x)=\frac{d}{dx}\left[3\right]+\frac{d}{dx}\left(9\cdot {(x+1)}^{-2}\right)\right]$

$=0+9\cdot \frac{d}{dx}\left({(x+1)}^{-2}\right)$

$=0+9\cdot \frac{d}{dx}\left({(x+1)}^{-2}\right)$

$=9\cdot \left(-2\right){(x+1)}^{-3}\frac{d}{dx}\left(x+1\right)=-18{(x+1)}^{-3}(1)=\frac{-18}{{(x+1)}^3}$;

$S\text{'}(2)=\frac{-18}{{(2+1)}^3}=-\frac{2}{3}\approx -.667$.

The equation $S(2)=4$

and $S\text{'}(2)=-.667$ tell us that on January $2$

the sales are $\$4000$

and are falling at the rate of $.667$ thousand dollars per day, or $\$667$ per day.

b. To estimate $S(3)$, we use $(3)$:

$S(3)\approx S(2)+S\text{'}(2)$

Thus, our estimate of the sales on January $3$ is $4000-667=\$3333$. To compare with the exact value of sale on January $3$, we compute $S(3)$

from the formula:

$S(3)=3+\frac{9}{{(3+1)}^2}=3+\frac{9}{16}=\frac{57}{16}=3.5625$ thousand dollars, or $\$3562.5$.

This is close to our estimate of $\$3333$.