The demand function for a product is p=24−3q where p is the price in dollars when q units are demanded. Find the level of production that maximizes the total revenue and determine the revenue.
Q=______units R_______
revenue=pg=24q-3q^2
d(revenue/dq)=0=24-6q
q=3
max revenut=24q-3q^2 compute that.
To find the level of production that maximizes the total revenue, we need to determine the quantity or level of production (q) that maximizes the product of price (p) and quantity (q) or the total revenue (R).
The demand function is given as p = 24 - 3q, where p is the price in dollars when q units are demanded.
The total revenue can be calculated by multiplying the price (p) by the quantity (q), so we have R = p * q.
Substituting the given demand function into the revenue equation, we get R = (24 - 3q) * q.
To find the level of production that maximizes revenue, we differentiate the revenue equation with respect to q and set the derivative equal to zero.
R = (24 - 3q) * q
dR/dq = 24q - 3q^2 (differentiating with respect to q)
Set dR/dq = 0 and solve for q:
24q - 3q^2 = 0
3q(8 - q) = 0
Setting each factor equal to zero, we get two possible solutions for q:
3q = 0 -> q = 0
8 - q = 0 -> q = 8
Since the quantity cannot be negative, the level of production that maximizes total revenue is q = 8 units.
To determine the revenue (R) at the level of production, substitute q = 8 into the revenue equation:
R = (24 - 3q) * q
R = (24 - 3 * 8) * 8
R = (24 - 24) * 8
R = 0 * 8
R = 0
Therefore, the level of production that maximizes the total revenue is q = 8 units and the revenue is $0.