q = 200 − 5 p
What is the (instantaneous) rate of change of revenue with respect to price when the price is $ 12?
At what price is the derivative of revenue with respect to price exactly zero?
To find the instantaneous rate of change of revenue with respect to price, we need to differentiate the revenue equation with respect to price.
Given that q = 200 − 5p, where q represents revenue and p represents price, we can differentiate the revenue equation using the power rule for differentiation, which states that the derivative of a constant multiplied by a variable raised to a power is equal to the constant times the power multiplied by the variable raised to the power minus one.
Differentiating q with respect to p, we get:
dq/dp = d(200 − 5p)/dp
= -5
This means that the instantaneous rate of change of revenue with respect to price is -5.
Therefore, no matter what the value of price is, the revenue is decreasing by $5 for each additional unit of price.
To find the price at which the derivative of revenue with respect to price is exactly zero, we need to set the derivative equal to zero and solve for p.
-5 = 0
Since there is no value of p that satisfies this equation, there is no price at which the derivative is exactly zero for this particular revenue equation.
To find the instantaneous rate of change of revenue with respect to price, we need to take the derivative of the revenue function with respect to price. Let's call the derivative dR/dp, where R represents revenue and p represents price.
Given that q = 200 - 5p, we can determine revenue by multiplying price and quantity:
R = q * p
Substituting the given expression for q, the revenue function becomes:
R = (200 - 5p) * p
To find the derivative of R with respect to p, we can apply the product rule:
dR/dp = (200 - 5p)*dp/dp + p*d(200 - 5p)/dp
Simplifying, we get:
dR/dp = -5p + 200 - 5p
Now, we can find the instantaneous rate of change of revenue by substituting the value p = 12 into the derivative equation:
dR/dp = -5(12) + 200 - 5(12)
= -60 + 200 - 60
= 80
Therefore, the instantaneous rate of change of revenue when the price is $12 is $80.
To find the price at which the derivative of revenue with respect to price is exactly zero, we set dR/dp equal to zero and solve for p:
-5p + 200 - 5p = 0
-10p + 200 = 0
-10p = -200
p = -200 / -10
p = 20
Therefore, the derivative of revenue with respect to price is zero when the price is $20.