The demand function of a good is given byP = 140 − 35Qwhere P denotes the price and Q the quantity.(a) Write down the revenue R as a function of the quantity Q.(b) Find the quantity Q that maximizes the revenue and find the maximumrevenue
Question
The demand function of a good is given byP = 140 − 35Qwhere P denotes the price and Q the quantity.(a) Write down the revenue R as a function of the quantity Q.(b) Find the quantity Q that maximizes the revenue and find the maximumrevenue
Solution
(a) The revenue (R) is given by the product of the price (P) and the quantity (Q). So, we can substitute the given demand function into this to get the revenue as a function of Q:
R = P * Q R = (140 - 35Q) * Q R = 140Q - 35Q^2
(b) To find the quantity (Q) that maximizes the revenue, we need to take the derivative of the revenue function with respect to Q, set it equal to zero, and solve for Q. This is because the maximum or minimum of a function occurs where its derivative is zero.
The derivative of R with respect to Q is:
dR/dQ = 140 - 70Q
Setting this equal to zero gives:
140 - 70Q = 0 70Q = 140 Q = 140 / 70 Q = 2
So, the quantity that maximizes the revenue is Q = 2.
To find the maximum revenue, we substitute Q = 2 into the revenue function:
R = 1402 - 352^2 R = 280 - 140 R = 140
So, the maximum revenue is 140.
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