A resident in a particular city pays a water bill of \( \$ 4.71 \) per month plus \( \$ 2.45 \) per 1000 gallons of water used (up to 4000 gallons). If this city has 544,000 customers that use less than 4000 gallons a month, find its monthly revenue function \( R(x) \), where the total number of gallons \( x \) is measured in thousands. \( R(x)=\square \) (Use integers or decimals for any numbers in the expression. Do not include the \( \$ \) symbol in your answer.)

To find the monthly revenue function \( R(x) \), let's break down the costs involved: 1. The fixed cost per customer is \( 4.71 \). 2. The variable cost is \( 2.45 \) for every 1000 gallons, so if \( x \) represents the thousands of gallons used, the contribution from the variable cost is \( 2.45x \) since \( x \) is in thousands. Now, we have: \[ R(x) = 544,000 \times (4.71 + 2.45x) \] Now, multiplying the fixed and variable components will give us: \[ R(x) = 544000 \times 4.71 + 544000 \times 2.45x \] Calculating \( 544000 \times 4.71 \): \[ 544000 \times 4.71 = 2,564,640 \] And \( 544000 \times 2.45 \): \[ 544000 \times 2.45 = 1,331,800 \] Putting it all together, we get: \[ R(x) = 2,564,640 + 1,331,800x \] Thus, the revenue function is: \[ R(x) = 2,564,640 + 1,331,800x \]