Question

Can someone walk me through this problem please?

Answer 1

Our cost is given by \(C(x)=54\sqrt{x}+\dfrac{x^2}{1000}\) where \(x\) is our quantity produced i.e. our production level. For (a), we merely let \(x=1050\):$$C(1050)=54\sqrt{1050}+\frac{1050^2}{1000}=270\sqrt{42}+\frac{2205}2$$

Answer 2

For our *average* cost, we divide our cost by the quantity:$$C(1050)/1050=\frac{9\sqrt6}{5\sqrt7}+\frac{21}{20}$$

Answer 3

Cost at 1050 will just be C(1050).

The average cost at 1050 will be \(\bf \frac{C(1050)-C(0)}{1050}\)

The marginal cost at 1050 is just C'(1050), so you'll need to find the derivative first.

The production level that will minimise the average cost will once again require you to find the minimum average rate of chance between x = 0 and some other x. Here the function being minimised is \(\bf \frac{C(x)-C(0)}{x}\).

The minimal average cost will just be the result of the above function being minised at the x-value that you get for the minimum average cost.

Answer 4

For our *marginal* cost, we compute the *derivative:$$C'(x)=\frac{54}{2\sqrt{x}}+\frac{2x}{1000}=\frac{27}{\sqrt{x}}+\frac{x}{500}\\C'(x)=\frac{27}{\sqrt{1050}}+\frac{1050}{500}=\frac{9\sqrt3}{5\sqrt{14}}+\frac{21}{10}$$

Answer 5

@oldrin.bataku I think you only need to guide him with the required steps. He can do the rest himself =]

Answer 6

Intuitively, this tells us this cost per *infinitesimal* increase in quantity, i.e. \(\delta C/\delta x\approx dC/dx\) for small \(\delta C,\delta x\)

Answer 7

To minimize average cost, we wish to minimize \(C(x)/x\) -- differentiate and find the relative extrema

Answer 8

@genius12 Thanks for your help!
@oldrin.bataku Thank you so much! I can take it from here! :D