the price p and the quantity x sold of a certain product obey the demand equation
p=(-1/3)x + 100
o is < or equal to x < or equal to 300
A) express revenue R as a function of x
b) what quantity x maximizes revenue.
A) To express revenue R as a function of x, we need to multiply the price p with the quantity x.
Revenue (R) = p * x
Substituting the given demand equation for p, we have:
R = ((-1/3)x + 100) * x
= (-1/3)x^2 + 100x
So, the revenue R can be expressed as a function of x as:
R(x) = (-1/3)x^2 + 100x
B) To find the quantity x that maximizes revenue, we need to find the vertex of the parabolic function R(x) = (-1/3)x^2 + 100x. The vertex of a parabola gives the maximum or minimum point of the function.
The x-coordinate of the vertex can be found using the formula: x = -b / 2a, where a and b are coefficients of the quadratic equation.
In this case, the equation is in the form R(x) = ax^2 + bx, where a = -1/3 and b = 100.
x = -b / 2a
= -100 / (2 * -1/3)
= -100 / (-2/3)
= -100 * (-3/2)
= 150
Therefore, the quantity x that maximizes revenue is x = 150.
To find the revenue function R(x) and the quantity that maximizes revenue, we need to understand the relationship between price, quantity, and revenue.
The revenue (R) is calculated as the product of the price (p) and the quantity (x): R = p * x.
A) Express revenue R as a function of x:
Given the demand equation p = (-1/3)x + 100, we can substitute this equation for p in the revenue equation:
R = ((-1/3)x + 100) * x.
Expanding the expression, we have:
R = (-1/3)x^2 + 100x.
Therefore, the revenue function R(x) is: R(x) = (-1/3)x^2 + 100x.
B) To find the quantity x that maximizes revenue, we can use different methods such as finding the critical points or graphing the function. Here, we'll use the critical points approach.
To find the critical points of R(x), we need to find where its derivative is zero or undefined. Taking the derivative of R(x) with respect to x, we get:
R'(x) = -2/3x + 100.
Now, let's solve R'(x) = 0 to find the critical points:
-2/3x + 100 = 0.
-2/3x = -100.
x = (-100) / (-2/3).
x = 150.
Therefore, when x = 150, the derivative R'(x) is equal to zero, indicating a possible maximum or minimum point.
To confirm that this is a maximum point, we can also check the second derivative, R''(x), where R''(x) > 0 indicates a maximum:
R''(x) = -2/3.
Since R''(x) is negative, this confirms that x = 150 is a maximum point.
Hence, the quantity x that maximizes revenue is 150.
Revenue=p*x
you finish..
Take the deriviative of R with respect to x, set to zero, solve for x.
so would the revenue be
(-1/3)x + 100 * 300
I don't understand- how would you multiply by x is its inequalities?
Also, what is the deriviative?