C(x) = 16x^(1/2) + x^2/1000
Find x to meet the minimum cost
Find the derivative, set equal to zero and solve for x
find C(x)
at x = 0
I think you have a typo probably.
As shown this function starts at zero and keeps getting bigger forever.
Did you leave a negative sign out?
No, there's no typo. That's what I can't figure this out. I put zero and it's wrong.
Beats me Carl.
To find the value of x that minimizes the cost, we need to find the value of x that corresponds to the minimum of the function C(x).
Step 1: Take the derivative of C(x) with respect to x.
C'(x) = (16 * (1/2)x^(-1/2)) + (2/1000)x
Simplifying, we get:
C'(x) = 8x^(-1/2) + (2/1000)x
Step 2: Set C'(x) equal to 0 and solve for x.
8x^(-1/2) + (2/1000)x = 0
We can simplify the equation by multiplying both sides by x^2:
8x + (2/1000)x^2 = 0
Step 3: Solve the quadratic equation.
To solve the quadratic equation, we can factor it as follows:
x(8 + (2/1000)x) = 0
So, either x = 0 or (8 + (2/1000)x) = 0
For x = 0, the cost C(x) would also be 0, but let's examine the second case:
8 + (2/1000)x = 0
(2/1000)x = -8
x = (-8)/(2/1000)
x = -4000
Since x cannot be negative in this context, we disregard x = -4000 as a valid solution.
Therefore, the value of x that minimizes the cost is x = 0.