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?
\[
p=-14 x+308
\]

Step 1 ：The price \(p\) and the quantity \(x\) sold of a small flat-screen television set obeys the demand equation \(p=-14x+308\).

Step 2 ：a) To find out how much should be charged for the television set if there are 50 television sets in stock, we substitute \(x=50\) into the equation. So, \(p=-14(50)+308\).

Step 3 ：Calculating the above expression, we get \(p=-700+308\).

Step 4 ：So, \(p=\boxed{-392}\).

Step 5 ：b) The revenue \(R\) is given by the product of the price \(p\) and the quantity \(x\), i.e., \(R=px\). Substituting the given equation into this, we get \(R=x(-14x+308)\), which simplifies to \(R=-14x^2+308x\).

Step 6 ：To find the quantity \(x\) that will maximize revenue, we take the derivative of the revenue function with respect to \(x\), set it equal to zero, and solve for \(x\). So, \(\frac{dR}{dx}=-28x+308=0\).

Step 7 ：Solving the above equation, we get \(x=\boxed{11}\).

Step 8 ：Substituting \(x=11\) into the revenue function, we get the maximum revenue \(R=-14(11)^2+308(11)\).

Step 9 ：Calculating the above expression, we get \(R=-14*121+308*11\).

Step 10 ：So, the maximum revenue is \(R=\boxed{1870}\).

Step 11 ：c) To find the price that should be charged in order to maximize revenue, we substitute \(x=11\) into the price function. So, \(p=-14(11)+308\).

Step 12 ：Calculating the above expression, we get \(p=-154+308\).

Step 13 ：So, the price that should be charged in order to maximize revenue is \(p=\boxed{154}\).