Verified answer

to find "x" where the cost is minimum, use the first derivative = 0

c'(x) = 110 - (1/x)(110)

0 = 110 - (1/x)(110)

(1/x)(110) = 110

110 = x.110

1 = x

to find the cost, input the (x=1) to the real equation

C(x)=600+110x−110ln(x)

C(1)=600+110(1)−110ln(1)

C(1) = 600 + 110 = 710

*note :

ln 1 = 0

derivative of (110)(ln x) is

U.V derivated -> U'V + V'U

so

U = 110

U' = 0

V = ln x

V' = 1/x

U'V + V'U ---> (0)(ln x) + (1/x)(110) = 0 + (1/x)(110) = (1/x)(110)

First, let the rate (C(x)) equal 0: 0 = (5x² - 21x + 3)/(3sqrt(x)) Now, considering that it's equal to zero, we most effective have to fear about the top considering the fact that for something to equal zero, the highest must be zero: 0 = 5x² - 21x + 3 Use the quadratic method to find the x values: x = 1/10 (21 ± sqrt[381]) x = 0.148078, four.05192 Now, let's determine to make sure none of these values make the long-established equation undefined: Subbing them into the normal equation, none of them make it undefined, so all the values are viable. Okay, we're practically carried out however x is the quantity of units and you can't have 0.14 models, so that doesn't work. The other root is close to four that will say it's four units in view that the rate could be very small (or marginal) when we plug it back into the equation.

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