The market demand for a certain product is $q$ units when the price charged to consumers is $p$ dollars, where $p = 1200 - 3q$. We need to find the total revenue that will be maximized.

Applied MathematicsOptimizationRevenue MaximizationCalculusDerivativesLinear Equations
2025/5/2

1. Problem Description

The market demand for a certain product is qq units when the price charged to consumers is pp dollars, where p=12003qp = 1200 - 3q. We need to find the total revenue that will be maximized.

2. Solution Steps

First, the total revenue RR is given by the product of the price pp and the quantity qq.
R=pqR = p \cdot q
Since p=12003qp = 1200 - 3q, we can substitute this into the revenue equation:
R=(12003q)qR = (1200 - 3q) \cdot q
R=1200q3q2R = 1200q - 3q^2
To maximize the revenue, we need to find the critical points by taking the derivative of RR with respect to qq and setting it equal to

0. $\frac{dR}{dq} = 1200 - 6q$

Setting the derivative to 0:
12006q=01200 - 6q = 0
6q=12006q = 1200
q=12006q = \frac{1200}{6}
q=200q = 200
Now, we need to find the price pp when q=200q = 200.
p=12003qp = 1200 - 3q
p=12003(200)p = 1200 - 3(200)
p=1200600p = 1200 - 600
p=600p = 600
Now we can find the maximum revenue RR by multiplying the price and quantity:
R=pqR = p \cdot q
R=600200R = 600 \cdot 200
R=120000R = 120000

3. Final Answer

The total revenue that will be maximized is $120,
0
0
0.

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