Answer: (i) Determine the Quadratic total revenue "R" equals "g" of "p":
Total revenue (R) is calculated by multiplying the quantity demanded (Q) by the price (p). Given the demand function Q equals F of p equals 30,000 minus 25p, we can express the total revenue function as:
R of p equals p times Q of p
Substituting the demand function into the total revenue function:
R of p equals p times (30,000 minus 25p)
R of p equals 30,000p minus 25p squared
So, the quadratic total revenue function is R of p equals 30,000p minus 25p squared.
(ii) What is the concavity of the function?
The concavity of the function can be determined by examining the coefficient of the quadratic term. If the coefficient of p squared term is negative, the function is concave down; if it's positive, the function is concave up. Here, the coefficient of p squared is negative 25, which means the function is concave down.
(iii) What is the R-intercept?
The R-intercept is the value of R of p when p equals 0. Substituting p equals 0 into the total revenue function:
R of 0 equals 30,000 times 0 minus 25 times 0 squared equals 0
So, the R-intercept is 0.
(iv) What does total revenue equal at a price of Rs. 60?
To find the total revenue at a price of Rs. 60, substitute p equals 60 into the total revenue function:
R of 60 equals 30,000 times 60 minus 25 times 60 squared
R of 60 equals 1,800,000 minus 90,000
R of 60 equals 1,710,000
So, the total revenue at a price of Rs. 60 is Rs. 1,710,000.
(v) How many units will be demanded at this price?
To find the quantity demanded at a price of Rs. 60, substitute p equals 60 into the demand function:
Q of 60 equals 30,000 minus 25 times 60
Q of 60 equals 30,000 minus 1500
Q of 60 equals 28,500
So, 28,500 units will be demanded at a price of Rs. 60.
(vi) At what price will total revenue be maximized?
Total revenue is maximized when the derivative of the total revenue function with respect to price is zero.
The derivative of R with respect to p equals 30,000 minus 50p equals 0
30,000 equals 50p
p equals 30,000 divided by 50 equals 600
So, total revenue will be maximized when the price is Rs. 600.
(vii) What is the maximum revenue?
To find the maximum revenue, substitute p equals 600 into the total revenue function:
R of 600 equals 30,000 times 600 minus 25 times 600 squared
R of 600 equals 18,000,000 minus 9,000,000
R of 600 equals 9,000,000
So, the maximum revenue is Rs. 9,000,000.
