Answer:
a)[tex]\bar{C}}(x)=0.0025x+80+\frac{10,000}{x}[/tex] b) 2000 cameras c) $90
Explanation:
a)The Average Cost function is:
[tex]\bar{C}}(x)=\frac{0.0025x^{2}+80x+10,000}{x}\\\bar{C}}(x)=0.0025x+80+\frac{10,000}{x}[/tex]
b) In order to find the level of production that results in the smallest average production cost. We need to proceed the calculations, finding out the 1st derivative of the Average Cost function. Then turn into an equality and find the level.
[tex]\bar{C}}(x)=0.0025x+80+\frac{10,000}{x}\\ \bar{C}}(x)=0.0025x+80+10,000x^{-1}\\ \bar{C}}(x)'=0.0025+0-10,000x^{-2}\\ \bar{C}}(x)'=0.0025-10,000x^{-2}\\0=0.0025-10,000x^{-2} \\\frac{10,000}{x^{2}}= \frac{25}{10000}\\\\25x^{2} =10^{8}\\x=\sqrt{\frac{10^{8}}{25}}\\ x=2000[/tex]
Test this finding with 2nd derivative test
[tex]\\ \\\bar{C}}(x)''=20,000x^{-3} \\ \bar{C}}(x)''=\frac{20,000}{2,000^{3}}\\ \bar{C}}(x)''=0.0000025[/tex]
So the 2nd derivative is ≈ 0. So that's right.
c) Find the level of production for which the average cost is equal to the marginal cost.
When Cannon reaches down the minimum average cost it is equalized to the marginal cost Average Cost= Marginal Cost.
[tex]\bar{C}}(x)=0.0025x+80+\frac{10000}{x} \\ 0.0025(2000)+80+\frac{10000}{2000}=\\ \bar{C}}(x)=5+80+5=90[/tex]
The level of production for which the average cost is equal to marginal cost is $90
In other words, the minimum average cost is $90 obtained producing 2000 cameras.
