Given The demand function of a product q=f (a)= 3600-15p find the maximizes total revenue and quanty sold

To find the quantity that maximizes total revenue, we need to use the demand function and the concept of revenue.

Total revenue (TR) is the product of the quantity sold (q) and the price per unit (p), which can be expressed as TR = q * p.

In this case, the demand function is given as q = f(a) = 3600-15p. We can rewrite this equation as p = (3600 - q)/15.

Substitute this value of p into the equation for total revenue:
TR = q * [(3600 - q)/15]

To find the quantity that maximizes total revenue, we need to take the derivative of the revenue function and set it equal to zero. Let's differentiate the equation with respect to q:

d(TR)/dq = [(3600 - q)/15] - [q/15]

Simplify:
d(TR)/dq = (3600 - 2q)/15

Setting this derivative equal to zero gives us:
(3600 - 2q)/15 = 0

Solve for q:
3600 - 2q = 0
2q = 3600
q = 3600/2
q = 1800

So, the quantity that maximizes total revenue (q) is 1800.

To find the maximum total revenue, substitute the value of q into the total revenue equation:
TR = q * [(3600 - q)/15]
TR = 1800 * [(3600 - 1800)/15]
TR = 1800 * (1800/15)
TR = 1800 * 120
TR = 216,000

Therefore, the quantity that maximizes total revenue is 1800, and the maximum total revenue is 216,000.