Solve the following problems The price p and the quantity x

Solve the following problems. The price p and the quantity x sold of a certain product obey the demand equation p = -13 x + 140, 0 lessthanorequalto x lessthanorequalto 420. What quantity x maximizes revenue? What is the maximum revenue?

Solution

p = (-x/3 + 140)

WE have x items..

So, total evenue is:
R = x * p

R = x(-x/3 + 140)

R = -x^2/3 + 140x

Now, for max revenue, we
just need to find x = -b/(2a) for this quadratic
where a = -1/3 and b = 140

We now have :
x = -140 / (2 * (-1/3))

x = -140 / (-2/3)

x = -140 * -3 / 2

x = 70*3

x = 210

So, the quantity that maximizes revenue is : 210

And when x = 210,
revenue is :
-x^2/3 + 140x
-210^2/3 + 140(210)
14700

So, answers :

Quantity that maximizes revenue is : 210
MAx revenue is : $14,700