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
80
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=-.16x+320
revenue is price * quantity, so
R(x) = x*p(x) = x(-.16x+320)
= -.16x^2 + 320x
Now just find the vertex of that parabola to locate maximum revenue.
a) To find the price of the television set when there are 80 sets in stock, we can substitute the value of x into the demand equation and solve for p.
Substituting x = 80 into the equation p = -0.16x + 320:
p = -0.16(80) + 320
p = -12.8 + 320
p = 307.2
Therefore, the price should be charged $307.2 for the television set when there are 80 sets in stock.
b) To find the quantity that will maximize revenue, we need to find the derivative of the revenue function and set it equal to zero.
The revenue function is given by R = px.
Differentiating R with respect to x:
dR/dx = dp/dx * x + p
Setting the derivative equal to zero:
0 = dp/dx * x + p
Since the demand equation is given as p = -0.16x + 320, we can substitute this into the equation:
0 = dp/dx * x + (-0.16x + 320)
Simplifying the equation:
0 = dp/dx * x - 0.16x + 320
Now we can solve for x by finding the value that satisfies this equation.
c) To find the price that should be charged to maximize revenue, we can substitute the value of x back into the demand equation.
Using the value of x from part b), substitute it into the demand equation p = -0.16x + 320:
p = -0.16(x value from part b) + 320
This will give us the price that should be charged to maximize revenue.
To answer these questions, we will use the given demand equation:
p = -0.16x + 320
Where p represents the price and x represents the quantity sold of the television set.
a) To find the price that should be charged for the television set when there are 80 sets in stock, we substitute x = 80 into the demand equation for p:
p = -0.16(80) + 320
p = -12.8 + 320
p ≈ 307.2
Therefore, the price that should be charged for the television set when there are 80 sets in stock is approximately $307.20.
b) To find the quantity that will maximize revenue and the maximum revenue, we need to find the maximum point on the revenue curve. Revenue is calculated by multiplying the price (p) by the quantity sold (x).
Revenue (R) = p * x
To find the maximum revenue, we can take the derivative of the revenue equation with respect to x and set it equal to zero to find the critical point.
dR/dx = 0
d/dx(p * x) = 0
dp/dx * x + p = 0 (since d/dx(x) = 1)
-0.16x + 320 = 0 (substituting the given demand equation)
-0.16x = -320
x = -320 / -0.16
x = 2000
The quantity that will maximize revenue is 2000.
To find the maximum revenue, substitute x = 2000 back into the demand equation to find the corresponding price (p), and then calculate the revenue (R) using the revenue equation:
p = -0.16(2000) + 320
p = -320 + 320
p = 0
R = p * x
R = 0 * 2000
R = 0
Therefore, the maximum revenue is $0.
c) To find the price that should be charged in order to maximize revenue, we can see from the previous calculation that the maximum revenue occurs when the price (p) is equal to 0. However, since we're talking about selling a television set, it doesn't make sense to reduce the price to $0. In practical terms, the price should be set higher to generate revenue.