Marginal Cost and Avg cost optimization problem

Question

I have this problem:

If $C(x) = 16,000 + 200x + 4x^{3/2}$, in dollars, find

1. the cost, average cost, and marginal cost at a production level of $1000$ units;
2. the production level that will minimize the average cost; and
3. the minimum average cost.

So for (i):

$$ C(x) = 16000 + 200x + 4x^{3/2} \implies C(1000) = 342491.11$$

So

Avg cost = $342491.11/1000 = 342.49$

Marg Cost function is given by $$ M(x) = C'(x) = 200 + 6x^{1/2} \implies M(1000) = 390.74 $$

(ii) $$ \begin{split} A(x) &= \frac{16000 + 200x + 4x^{3/2}}{x}\\ A'(x) &= \frac{x \cdot (200 + 6x^{1/2} - (16000 - 200x + 4x^{3/2}}{x^2}\\ &= \frac{200 + 6x^{\frac{3}{2}} - (16000 - 200x - 4x^{3/2}}{x^2}\\ &= \frac{2x^{3/2} - 16000}{x^2} \\ &= 0 \end{split} $$

when $2x^{\frac{3}{2}} = 16000 \implies x^{3/2} = 8000$

So after taking the cube root of 8000 and squaring it, $x = 400$

Did I make any mistakes?

(iii) 320

Answer 1

all are correct but (ii) may seem more direct if you divide the fraction to get $$ A(x) = 16000 x^{-1} + 200 + 4x^{1/2} $$ so $$ A'(x) = 2x^{-1/2} - 16000x^{-2} $$ thus $A'(x) = 0$ yields $8000 = x^{3/2}$ so $x = 400$.

Answer 2

your first derivative can simplified to $$A'(x)=2\,{\frac { \left( \sqrt {x}-20 \right) \left( x+20\,\sqrt {x}+400 \right) }{{x}^{2}}} $$ and $$A'(x)=0$$ if $x=400$ as you stated.