Suppose the function p= -1/8 x + 100 (0 grtr&= x ~12) relates the selling price p of an item to the quantity x that is sold. Assume that p is in dollars. What is the maximum revenue possible in this situation?
Suppose the function p= -1/8 x + 100
(0 grtr&= x ~12) relates the selling
price p of an item to the quantity x
that is sold. Assume that p is in
dollars. What is the maximum
revenue possible in this situation?
You should use the formula for revenue R(x)=p*q.
In this case q is x. So you do R(x)= -1/8 x + 100(x). This results in R(x)= -1/8x^2 + 100x.
To find the max, set this to 0 and complete the square:
0=-1/8(x^2-800x)
0=-1/8(x-400)^2+20000
20,000 is your answer
To find the maximum revenue in this situation, we need to determine the quantity that will result in the maximum value of the product of the selling price and quantity.
The revenue is given by the equation R = p * x, where R is the revenue, p is the selling price, and x is the quantity sold.
In this case, the selling price function is given as p = -1/8x + 100.
Substituting the value of p into the revenue equation, we have:
R = (-1/8x + 100) * x
Expanding the equation, we get:
R = (-1/8)x^2 + 100x
Now, we need to find the value of x that maximizes this function. This can be done by finding the vertex of the parabolic function, which occurs at x = -b / (2a), where a and b are the coefficients of the quadratic equation.
In this case, a = -1/8 and b = 100. Substituting these values into the vertex formula, we have:
x = -(100) / (2 * (-1/8))
x = -800 / (-1/4)
x = -800 * (-4/1)
x = 3200
Therefore, the quantity that maximizes the revenue is x = 3200.
To find the maximum revenue, we can substitute this value back into the revenue function:
R = (-1/8 * 3200^2) + 100 * 3200
R = (-1/8 * 10240000) + 320000
R = -1280000 + 320000
R = -960000
Therefore, the maximum revenue possible in this situation is $960,000.
To find the maximum revenue in this situation, we need to understand that revenue is calculated by multiplying the selling price (p) by the quantity sold (x).
The given function is:
p = -1/8x + 100
To calculate the revenue, we will multiply p by x:
Revenue (R) = p * x
Since we want to find the maximum revenue, we need to determine the maximum value of R.
We can rewrite the equation for p in terms of x:
p = -1/8x + 100
Multiply both sides of the equation by x:
px = -1/8x^2 + 100x
Now, substitute the expression for p in terms of x into the equation for revenue:
R = (-1/8x + 100) * x
R = -1/8x^2 + 100x
To find the maximum revenue, we need to determine the maximum point on the graph of R.
The graph of R is a parabolic curve that opens downward since the coefficient of x^2 is negative. The vertex of the parabola represents the maximum point.
The formula for the x-coordinate of the vertex is given by:
x = -b / (2a)
In our case, the equation of R is:
R = -1/8x^2 + 100x
Comparing it with the general equation for a parabola in the form of ax^2 + bx + c, we get:
a = -1/8
b = 100
Now, substitute the values of a and b into the formula for the x-coordinate of the vertex:
x = -100 / (2 * (-1/8))
x = -100 / (-1/4)
x = -100 * -4/1
x = 400
So, x = 400 is the x-coordinate of the vertex, which represents the quantity (x) that will yield the maximum revenue.
Now, substitute this value of x back into the equation for R to find the corresponding maximum revenue:
R = -1/8x^2 + 100x
R = -1/8(400)^2 + 100(400)
R = -1/8(160,000) + 40,000
R = -20,000 + 40,000
R = 20,000
Therefore, the maximum revenue possible in this situation is $20,000.