Question

section 2.12: implicit differentiation and related rates (homework)
score: 50/90 answered: 5/9
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question 6
0/10 pts 4 99 details
find the rate of change of total profit, in dollars,
with respect to time where ( r(x)=90x - 0.5x^{2} )
and ( c(x)=30x + 6 ), when ( x = 45 ) and
( \frac{dx}{dt}=90 ).
question help: video

Explanation:

Step1: Define profit function

Profit \( P(x)=R(x)-C(x)\). Substitute \( R(x) = 90x - 0.5x^{2}\) and \( C(x)=30x + 6\).
\( P(x)=(90x - 0.5x^{2})-(30x + 6)=60x-0.5x^{2}-6\)

Step2: Differentiate profit function with respect to \(t\)

Using the chain - rule \(\frac{dP}{dt}=\frac{dP}{dx}\cdot\frac{dx}{dt}\).
Differentiate \(P(x)\) with respect to \(x\): \(\frac{dP}{dx}=60 - x\)

Step3: Substitute values

Given \(x = 45\) and \(\frac{dx}{dt}=90\)
\(\frac{dP}{dt}=(60 - x)\cdot\frac{dx}{dt}\)
Substitute \(x = 45\) into \(\frac{dP}{dx}\): \(\frac{dP}{dx}=60-45 = 15\)
Then \(\frac{dP}{dt}=15\times90\)

Answer:

\(1350\)