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.

1. Problem Description

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.

2. Solution Steps

First, the total revenue $R$ is given by the product of the price $p$ and the quantity $q$.

$R = p \cdot q$

Since $p = 1200 - 3q$, we can substitute this into the revenue equation:

$R = (1200 - 3q) \cdot q$

$R = 1200q - 3q^2$

To maximize the revenue, we need to find the critical points by taking the derivative of $R$ with respect to $q$ and setting it equal to

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

Setting the derivative to 0:

$1200 - 6q = 0$

$6q = 1200$

$q = \frac{1200}{6}$

$q = 200$

Now, we need to find the price $p$ when $q = 200$.

$p = 1200 - 3q$

$p = 1200 - 3(200)$

$p = 1200 - 600$

$p = 600$

Now we can find the maximum revenue $R$ by multiplying the price and quantity:

$R = p \cdot q$

$R = 600 \cdot 200$

$R = 120000$

3. Final Answer

The total revenue that will be maximized is $120,

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