The price $p$ and the quantity $x$ sold of a small flat-screen television set obeys the demand equation below. a) How much should be charged for the television set if there are 50 television sets in stock? b) What quantity $x$ will maximize revenue? What is the maximum revenue? c) What price should be charged in order to maximize revenue? \[ p=-.15 x+450 \] a) What should be charged for the units in stock? $\$ 442.5$ (Round to the nearest cent as needed.) b) What quantity will maximize the revenue? television sets

Step 1 ：Given the demand equation \(p=-0.15x+450\), we are asked to find the price that should be charged for the television set if there are 50 television sets in stock.

Step 2 ：Substitute \(x=50\) into the equation to find the price \(p\).

Step 3 ：\(p=-0.15(50)+450=442.5\). So, the price that should be charged for the television set if there are 50 television sets in stock is \(\boxed{\$442.50}\).

Step 4 ：To find the quantity \(x\) that will maximize revenue, we need to find the derivative of the revenue function \(R=px\) with respect to \(x\), set it to zero, and solve for \(x\).

Step 5 ：The revenue function is \(R=x(450 - 0.15x)\).

Step 6 ：The derivative of the revenue function is \(R'=450 - 0.3x\).

Step 7 ：Setting \(R'=0\), we get \(450 - 0.3x = 0\). Solving for \(x\), we get \(x=1500\).

Step 8 ：Substitute \(x=1500\) into the revenue function to find the maximum revenue \(R\).

Step 9 ：\(R=1500(450 - 0.15(1500))=337500\). So, the quantity \(x\) that will maximize revenue is 1500 television sets and the maximum revenue is \(\boxed{\$337,500}\).