Demand Equation:
The price p, in dollars, and the quantity x sold of a certain product obey the demand equation
x = -5p + 100, 0 (less than or equal to) p (less than or equal to) 20
a.) Express the revenue R as a funtion of x.
b.) What is the revenue if 15 units are sold?
c.) What quantity x maximizes revenue? What is the maximum revenue?
d.) What price should the company charge to maximize revenue?
Revenue= P*x
but p= 1/5 x +20 check that, then substitute to find Revenue as a function of x.
Maximizing Revenue. In calculus, there is a very elegant way. In algebra, we want find when Revenue is max.
Revenue= P*x
but p= 1/5 x +20 check that, then substitute to find Revenue as a function of x.
Maximizing Revenue. In calculus, there is a very elegant way. In algebra, we want find when Revenue is max.
The revenue function for a bicycle shop is given by R(x) = x · p(x) dollars where x is the number of units sold and p(x) = 100 − 0.5x is the unit price. Find the maximum revenue.
To express the revenue R as a function of x, we can substitute the value of p from the demand equation into the revenue equation.
Revenue = p * x
From the demand equation, we have p = (1/5)x + 20.
Substituting this into the revenue equation, we get:
Revenue = ((1/5)x + 20) * x
= (1/5)x^2 + 20x
So the revenue R is a function of x given by R = (1/5)x^2 + 20x.
To find the revenue if 15 units are sold, we can substitute x = 15 into the revenue equation:
Revenue = (1/5)(15)^2 + 20(15)
= (1/5)(225) + 300
= 45 + 300
= 345 dollars
Therefore, the revenue if 15 units are sold is 345 dollars.
To find the quantity x that maximizes revenue, we need to find the maximum value of the revenue function R = (1/5)x^2 + 20x. This can be done by finding the vertex of the parabola represented by the function.
The x-coordinate of the vertex can be found using the formula x = -b/2a, where a = 1/5 and b = 20. Substituting these values, we get:
x = -(20)/(2(1/5))
= -20 / (2/5)
= -20 * (5/2)
= -50
So the quantity x that maximizes revenue is -50 units. However, since the quantity of units sold cannot be negative, we disregard this result.
Next, we should consider the endpoints of the domain of the demand equation, which are 0 and 20.
Substituting 0 into the revenue equation, we get:
Revenue = (1/5)(0)^2 + 20(0)
= 0 + 0
= 0 dollars
Substituting 20 into the revenue equation, we get:
Revenue = (1/5)(20)^2 + 20(20)
= (1/5)(400) + 400
= 80 + 400
= 480 dollars
Therefore, the quantity x that maximizes revenue is 20, and the maximum revenue is 480 dollars.
To find the price that the company should charge to maximize revenue, we can substitute x = 20 into the demand equation:
x = -5p + 100
20 = -5p + 100
-5p = -80
p = 16
Therefore, the price the company should charge to maximize revenue is 16 dollars.
To express revenue R as a function of x, we use the equation R = p * x. Since the demand equation tells us that p = (1/5)x + 20, we can substitute this expression for p into the revenue equation:
R = ((1/5)x + 20) * x
Simplifying this equation gives us the revenue as a function of x.
To find the revenue when 15 units are sold, we substitute x = 15 into the revenue equation:
R = ((1/5)*15 + 20) * 15
Calculate the expression to find the revenue.
To find the quantity x that maximizes revenue, we want to find the value of x that makes the revenue function R as large as possible. In calculus, we find this maximum by taking the derivative of the revenue function with respect to x and setting it equal to zero. However, if you prefer an algebraic approach, you can also find the maximum by substituting different values of x into the revenue equation and finding the largest corresponding revenue.
To find the maximum revenue, you would substitute the value of x that maximizes revenue into the revenue equation.
To find the price the company should charge to maximize revenue, we need to find the corresponding value of p when x is at its maximum. This can be done by substituting the value of x that maximizes revenue into the demand equation and solving for p.