Respuesta :
Answer:
a) [tex]\frac{74}{10025}[/tex]
b) [tex]\frac{3x-2}{x(8x+1)}[/tex]
c) [tex]\frac{-24x^2+32x-2}{(8x^2+x)^2}[/tex]
Step-by-step explanation:
For total cost function [tex]c(x)[/tex], average cost is given by [tex]\frac{c(x)}{x}[/tex] i.e., total cost divided by number of units produced.
Marginal average cost function refers to derivative of the average cost function i.e., [tex]\left ( \frac{c(x)}{x} \right )'[/tex]
Given:[tex]c(x)=\frac{3x-2}{8x+1}[/tex]
Average cost = [tex]\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}[/tex]
a)
At x = 50 units,
[tex]\frac{c(50)}{50}=\frac{150-2}{50(400+1)}=\frac{148}{50(401)}=\frac{74}{10025}[/tex]
b)
Average cost = [tex]\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}[/tex]
c)
Marginal average cost:
Differentiate average cost with respect to [tex]x[/tex]
Take [tex]f=3x-2\,,\,g=8x^2+x[/tex]
using quotient rule, [tex]\left ( \frac{f}{g} \right )'=\frac{f'g-fg'}{g^2}[/tex]
Therefore,
[tex]\left ( \frac{c(x)}{x} \right )'=\left ( \frac{3x-2}{8x^2+x} \right )'\\=\left ( \frac{3(8x^2+x)-(16x+1)(3x-2)}{(8x^2+x)^2} \right )\\=\frac{24x^2+3x-48x^2-3x+32x+2}{(8x^2+x)^2}\\=\frac{-24x^2+32x-2}{(8x^2+x)^2}[/tex]
1) The average cost for the production at 50 units and x units respectively are;
A) A(50) = 74/10025
B) A(x) = (3x - 2)/(8x² + x)
2) The marginal average cost function is; M(x) = [tex]\frac{-24x^{2} + 32x + 2}{x^{2}(8x^{2} + 1)^{2} }[/tex]
We are given Total cost function as;
C(x) = (3x - 2)/(8x + 1)
The formula for the average cost is;
A(x) = (C(x))/x
where x is the units of the product
Thus;
A) At x = 50 units;
A(50) = [(3(50) - 2)/(8(50) + 1)]/50
A(50) = (148/401)/50
A(50) = 74/10025
B) At x units;
A(x) = [(3x - 2)/(8x + 1)]/x
A(x) = (3x - 2)/[x(8x + 1)]
A(x) = (3x - 2)/(8x² + x)
C) Formula for average marginal cost function is;
M(x) = [(C(x))/x]'
(C(x))/x = (3x - 2)/(8x² + x)
Thus;
M(x) = ['(3x - 2)/(8x² + x)]'
From online differentiation calculator, we have;
M(x) = [tex]\frac{-24x^{2} + 32x + 2}{x^{2}(8x^{2} + 1)^{2} }[/tex]
Read more about marginal cost function at; https://brainly.com/question/11689872
