Demand Equation. The price(in dollars),and the quantity x sold of a certain product, obey the demand equation. I can't figure out what price the company should charge to maximize revenue
p=-1/10x+150
R=p*x
R=((-1/10)*x+150)*x
R=(-1/10)*x^2+150x
y=ax^2+b*x+c
a=-1/10
b=150
c=0
-2 b/2a=150/2*(-1/10)
-b/2a=150/1/5
-b/2a=750
R max=(-1/10)*x^2+150x
R max=(-1/10)*750^2+150*750
R max=-562500/10+112500
R max=-56250+112500
R max=56250
well, recall what you were told:
The price(in dollars) p,and the quantity sold x obey the demand equation
p=-1/10x+150
You found that maximum revenue R occurs at x=750, so
p = -75+150 = 75
To determine the price that the company should charge to maximize revenue, we need to find the value of x (quantity) that corresponds to the maximum value of R (revenue).
Given the demand equation: p = -1/10x + 150
The revenue equation is given by: R = p * x
Substituting the value of p from the demand equation into the revenue equation, we get:
R = (-1/10x + 150) * x
Expanding the equation, we have:
R = (-1/10)x^2 + 150x
This is a quadratic equation in terms of x. To find the value of x that maximizes R, we can use the formula x = -b / 2a, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c.
In this case, a = -1/10 and b = 150. Therefore, substituting these values into the formula:
x = -(150) / 2(-1/10)
Simplifying, we have:
x = -(150) / (-2/10)
x = (150) / (2/10)
x = (150) * (10/2)
x = 1500 / 2
x = 750
So, the company should charge a price that will result in a quantity of 750 units sold in order to maximize revenue.
To find the maximum revenue, substitute the value of x into the revenue equation:
R = (-1/10)(750)^2 + 150(750)
R = -562500/10 + 112500
R = -56250 + 112500
R = 56250
Therefore, the maximum revenue is $56,250.