Mathematics, 14.01.2020 22:31 justachelseafan

Minimum average cost the cost of producing units of a product is modeled by
c = 200 + 75x - 300 in x, x > 1.
(a) find the average cost function c.
(b) find the minimum average cost analytically. use a graphing utility to conform your rest.

Answers: 3

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Answer from: baileyanne9389

Answer and explanation:

A)The average cost function C is obtain by dividing the cost function c by the number of products x:

$C(x)=\frac{200}{x} +75-\frac{300ln(x)}{x}$

B) the minimum average cost can be obtained by obtaining the minimum of the function deriving the function and equalizing it to zero:

$\frac{dC(x)}{dx} =0\\\frac{d}{dx} (\frac{200}{x} +75-\frac{300ln(x)}{x})=0\\\frac{-200}{x^2} +\frac{300ln(x)}{x^2}-\frac{300}{x^2}=0\\\frac{1}{x^2}(-500+300ln(x))=0$

$\frac{1}{x^2}=0$ ↔x=0 but x>0

$(-500+300ln(x))=0\\ln(x)=5/3\\X=5.29$

But x is an integer so the minimum is found in x=5 so:

C(5)=18.43

Minimum average cost the cost of producing units of a product is modeled by c = 200 + 75x - 300 in x

Answer from: cxttiemsp021

a) $C_{av}(x)=\frac{200}{x}+75-300\frac{ln x}{x}$

b) $C_{av min}=75-\frac{300}{e^{5/3}}=18.34$

Step-by-step explanation:

a) In order to solve this part of the problem we must remember that the average cost function is found by using the following formula:

$C_{av}=\frac{C(x)}{x}$

so all we need to do is to divide the cost function into x, so we get:

$C_{av}=\frac{200+75x-300 ln x}{x}$

which yields:

$C_{av}(x)=\frac{200}{x}+75-\frac{ln x}{x}$

b) In order to find the minimum average cost analytically, we must use the first derivative to find the local minima, so we proceed and find the derivative of the average cost function. We can rewrite the function so it is easier for us to find:

$C_{av}=200x^{-1}+75-\frac{lnx}{x}$

we will need to find the derivative of a fraction so we will be using the quotient rule to do so, the quotient rule goes as follows:

$(\frac{p}{q})'=\frac{qp'-pq'}{q^{2}}$

in this case we get that:

p=lnx

$p'=\frac{1}{x}$

q=x

q'=1

so when using the formula we get that the derivative of $300\frac{ln x}{x}$ is:

$300\frac{\frac{x}{x}-ln x}{x^{2}}$

which simplifies to:

$300\frac{1-ln x}{x^{2}}$

so the whole derivative of the average cost function turns out to be:

$C'_{av}(x)=-200x^{-2}-300\frac{1-ln x}{x^{2}}$

Which can be rewritten as:

$C'_{av}(x)=-\frac{200}{x^{2}}-300\frac{1-ln x}{x^{2}}$

so now we can set it equal to zero to solve for x, so we get that:

$-\frac{200}{x^{2}}-300\frac{1-ln x}{x^{2}}=0$

So now we multiply both sides of the equation by $x^{2}$ .

so we get:

-200-300+300ln x=0

300lnx=500

so:

$ln x=\frac{5}{3}$

so the answer is:

$x=e^{5/3}$

Now we need to determine if this is a local max or min. We can do so by using the second derivative test, so we take the second derivative of the average cost function:

$C'(x)_{av}=-200x^{-2}-300x^{-2}+300\frac{ln x}{x^{2}}$