Answer :
We start with the relationship between the weekly sales [tex]\( s \)[/tex] and the weekly advertising costs [tex]\( x \)[/tex] given by
[tex]$$
s=60000-320000e^{-0.0009x}.
$$[/tex]
We are told that the advertising costs are currently [tex]\( x=2000 \)[/tex] dollars and that the advertising costs are increasing at a rate of
[tex]$$
\frac{dx}{dt}=300 \text{ dollars per week}.
$$[/tex]
Our goal is to find the current rate of change of sales [tex]\( \frac{ds}{dt} \)[/tex], which we will obtain by using the chain rule:
[tex]$$
\frac{ds}{dt}=\frac{ds}{dx}\cdot\frac{dx}{dt}.
$$[/tex]
### Step 1. Find [tex]\(\frac{ds}{dx}\)[/tex]
Differentiate [tex]\( s \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]$$
\frac{ds}{dx}=\frac{d}{dx}\left[60000-320000e^{-0.0009x}\right].
$$[/tex]
The derivative of the constant [tex]\( 60000 \)[/tex] is [tex]\( 0 \)[/tex]. For the second term, apply the chain rule. The derivative of [tex]\( e^{-0.0009x} \)[/tex] is
[tex]$$
\frac{d}{dx}\left(e^{-0.0009x}\right)=-0.0009e^{-0.0009x}.
$$[/tex]
Thus,
[tex]$$
\frac{ds}{dx}=-320000\left(-0.0009e^{-0.0009x}\right)=320000\cdot0.0009e^{-0.0009x}.
$$[/tex]
Note that
[tex]$$
320000\cdot0.0009=288,
$$[/tex]
so we can write
[tex]$$
\frac{ds}{dx}=288e^{-0.0009x}.
$$[/tex]
### Step 2. Evaluate [tex]\(\frac{ds}{dx}\)[/tex] at [tex]\( x=2000 \)[/tex]
Substitute [tex]\( x=2000 \)[/tex] into the expression:
[tex]$$
\frac{ds}{dx}=288e^{-0.0009(2000)}.
$$[/tex]
Calculate the exponent:
[tex]$$
0.0009\times2000=1.8,
$$[/tex]
which gives
[tex]$$
\frac{ds}{dx}=288e^{-1.8}.
$$[/tex]
Numerically, [tex]\( e^{-1.8}\approx0.1652988882 \)[/tex]. Therefore,
[tex]$$
\frac{ds}{dx}\approx288\times0.1652988882\approx47.60608.
$$[/tex]
### Step 3. Compute [tex]\(\frac{ds}{dt}\)[/tex]
Now, use the chain rule:
[tex]$$
\frac{ds}{dt}=\frac{ds}{dx}\cdot\frac{dx}{dt}.
$$[/tex]
Substitute the computed values:
[tex]$$
\frac{ds}{dt}\approx47.60608\times300.
$$[/tex]
Multiplying these,
[tex]$$
\frac{ds}{dt}\approx14281.82394.
$$[/tex]
### Final Answer
The current rate of change of sales is approximately
[tex]$$
\boxed{14281.82 \text{ dollars per week}}.
$$[/tex]
[tex]$$
s=60000-320000e^{-0.0009x}.
$$[/tex]
We are told that the advertising costs are currently [tex]\( x=2000 \)[/tex] dollars and that the advertising costs are increasing at a rate of
[tex]$$
\frac{dx}{dt}=300 \text{ dollars per week}.
$$[/tex]
Our goal is to find the current rate of change of sales [tex]\( \frac{ds}{dt} \)[/tex], which we will obtain by using the chain rule:
[tex]$$
\frac{ds}{dt}=\frac{ds}{dx}\cdot\frac{dx}{dt}.
$$[/tex]
### Step 1. Find [tex]\(\frac{ds}{dx}\)[/tex]
Differentiate [tex]\( s \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]$$
\frac{ds}{dx}=\frac{d}{dx}\left[60000-320000e^{-0.0009x}\right].
$$[/tex]
The derivative of the constant [tex]\( 60000 \)[/tex] is [tex]\( 0 \)[/tex]. For the second term, apply the chain rule. The derivative of [tex]\( e^{-0.0009x} \)[/tex] is
[tex]$$
\frac{d}{dx}\left(e^{-0.0009x}\right)=-0.0009e^{-0.0009x}.
$$[/tex]
Thus,
[tex]$$
\frac{ds}{dx}=-320000\left(-0.0009e^{-0.0009x}\right)=320000\cdot0.0009e^{-0.0009x}.
$$[/tex]
Note that
[tex]$$
320000\cdot0.0009=288,
$$[/tex]
so we can write
[tex]$$
\frac{ds}{dx}=288e^{-0.0009x}.
$$[/tex]
### Step 2. Evaluate [tex]\(\frac{ds}{dx}\)[/tex] at [tex]\( x=2000 \)[/tex]
Substitute [tex]\( x=2000 \)[/tex] into the expression:
[tex]$$
\frac{ds}{dx}=288e^{-0.0009(2000)}.
$$[/tex]
Calculate the exponent:
[tex]$$
0.0009\times2000=1.8,
$$[/tex]
which gives
[tex]$$
\frac{ds}{dx}=288e^{-1.8}.
$$[/tex]
Numerically, [tex]\( e^{-1.8}\approx0.1652988882 \)[/tex]. Therefore,
[tex]$$
\frac{ds}{dx}\approx288\times0.1652988882\approx47.60608.
$$[/tex]
### Step 3. Compute [tex]\(\frac{ds}{dt}\)[/tex]
Now, use the chain rule:
[tex]$$
\frac{ds}{dt}=\frac{ds}{dx}\cdot\frac{dx}{dt}.
$$[/tex]
Substitute the computed values:
[tex]$$
\frac{ds}{dt}\approx47.60608\times300.
$$[/tex]
Multiplying these,
[tex]$$
\frac{ds}{dt}\approx14281.82394.
$$[/tex]
### Final Answer
The current rate of change of sales is approximately
[tex]$$
\boxed{14281.82 \text{ dollars per week}}.
$$[/tex]
