A total revenue function is given by R(x) = 1000(x^2 - 0.1x)^1/2 , where R(x) is the total revenue, in thousands of dollars, from the sale of x items. Find the rate at which total revenue is changing when 20 items have been sold.
So far my answer is
R'(x) = 1000
Coming from R'(x) = 500 ((x^2) - 0.1x)^-.5 (2x - 0.1)
That is correct.
To find the rate at which total revenue is changing when 20 items have been sold, we need to find the derivative of the total revenue function R(x) with respect to x.
The total revenue function is given by:
R(x) = 1000(x^2 - 0.1x)^1/2
To find the derivative, we can use the chain rule. The chain rule states that if we have a function h(u) = f(g(u)), then the derivative of h(u) with respect to u is given by h'(u) = f'(g(u)) * g'(u).
For our function R(x) = 1000(x^2 - 0.1x)^1/2, let's define f(u) = 1000u^1/2 and g(x) = x^2 - 0.1x. Notice that R(x) = f(g(x)).
First, let's find the derivative of f(u):
f'(u) = (1000 * 1/2)(u^(-1/2))
f'(u) = 500u^(-1/2)
f'(u) = 500/u^1/2
Next, let's find the derivative of g(x):
g'(x) = 2x - 0.1
Now, let's apply the chain rule to find the derivative of R(x) with respect to x:
R'(x) = f'(g(x)) * g'(x)
R'(x) = 500/(x^2 - 0.1x)^1/2 * (2x - 0.1)
Notice that your partial derivative R'(x) = 500(x^2 - 0.1x)^-0.5(2x - 0.1).
Now, to find the rate at which total revenue is changing when 20 items have been sold, we substitute x = 20 into R'(x):
R'(20) = 500(20^2 - 0.1(20))^-0.5(2 * 20 - 0.1)
Evaluating this expression will give you the rate at which total revenue is changing when 20 items have been sold.
To find the rate at which total revenue is changing when 20 items have been sold, we need to find the derivative of the total revenue function, R(x), with respect to x, and then substitute x = 20.
Given: R(x) = 1000(x^2 - 0.1x)^(1/2)
To find R'(x), we can use the chain rule.
First, let's rewrite the equation as R(x) = 1000(x^2 - 0.1x)^(1/2) = 1000u^0.5, where u = (x^2 - 0.1x).
Now, we can find the derivative of R(x) with respect to u, and then multiply it by the derivative of u with respect to x.
dR/du = 0.5 * 1000 * u^(-0.5) = 500(u^(-0.5))
Next, we find the derivative of u with respect to x:
du/dx = d/dx (x^2 - 0.1x) = 2x - 0.1
Now, we multiply dR/du by du/dx to get the derivative of R(x) with respect to x:
R'(x) = 500(u^(-0.5)) * (2x - 0.1)
Substituting u = (x^2 - 0.1x) back into the equation, we have:
R'(x) = 500((x^2 - 0.1x)^(-0.5))(2x - 0.1)
So, your answer is correct:
R'(x) = 500((x^2 - 0.1x)^(-0.5))(2x - 0.1)
Now, we can find the rate at which total revenue is changing when 20 items have been sold by substituting x = 20 into the equation:
R'(20) = 500((20^2 - 0.1 * 20)^(-0.5))(2 * 20 - 0.1)
Simplifying this expression will give you the exact value of the rate at which total revenue is changing when 20 items have been sold.