Question

for the given cost function C(x)=16sqrt(x)+(x^2/3375) find a) the cost at the production level 2000. b) the average cost at the production level 2000. c) the marginal cost at the production level 2000. d) the production level that will minimize the average cost. e) the minimal average cost

Answer 1

\[C(x) = 16\sqrt{x} + \frac{x^2}{3375}\]
a.) Find \(C(2000)\). For this part just replace the variable \(x\) with the number \(2000\).

Answer 2

i tried doing that but it says my answer is wrong

Answer 3

What are you getting?

Answer 4

320* sqrt(5) +(3200/27)

Answer 5

wow...ok:) I would have thought you could just use the calculator for this (application problem). I was thinking the decimal answer of about $1,900.73

Answer 6

ohh thats right! can you help me with the other parts

Answer 7

Sure.

Answer 8

b.) Average cost = \(\overline{C(x)} = \dfrac{C(x)}{x} \)
\[= \dfrac{16\sqrt{x} + \dfrac{x^2}{3375}}{x}\]
Best to simplify this a bit.
\[\dfrac{16\sqrt{x} }{x}+ \dfrac{x^2}{3375x}\]
Ok so far?

Answer 9

i got part b

Answer 10

yeah i got that answer i need help with c d and e

Answer 11

Ok.. so I simplified part b to
\[\overline{C(x)} = \frac{16}{\sqrt{x}} + \frac{x}{3375}\]

c.) Marginal cost = derivative of \(\overline{C(x)}\)

Answer 12

how can ypu get the derivit for c.) ?

Answer 13

crap..
marginal cost = derivative of cost
marginal average cost = derivative of average cost ...

Answer 14

\[C(x) = 16\sqrt{x} + \frac{x^2}{3375}\]
\[\large C(x) = 16x^{1/2}+ \frac{1}{3375}x^2\]
use power rule:
\[\large C'(x) = 16\left( \frac{1}{2}x^{-1/2}\right) + \frac{1}{3375}(2x)\]

Answer 15

so is that the answer to c?

Answer 16

Simplify a bet and plug in 2000.

Answer 17

Should be about $1.36 per item
http://www.wolframalpha.com/input/?i=deravitive+of+16*sqrt%28x%29+%2B+%28x%29^2%2F3375+at+x+%3D+2000

Answer 18

its coming out wrong

Answer 19

is the 1.36 for part c?

Answer 20

yes

Answer 21

yeah its coming out wrong

Answer 22

but ill skip that for now, do you now how to do part d and e?

Answer 23

are they maybe asking for marginal average cost???

Answer 24

the question is asking for marginal cost but ill try entering marginal average cost
how do i find that?

Answer 25

You would find the derivative of average cost using power rule. It's a very small #. Rounding it would be $0.00!!
http://www.wolframalpha.com/input/?i=derivative%28+%2816*sqrt%28x%29+%2B+%28x%29^2%2F3375%29%2Fx%2Cx%29+at+x+%3D+2000

Answer 26

its still coming out wrong :/

Answer 27

:( not sure then.

Answer 28

can i do the other parts skipping this part

Answer 29

d) the production level that will minimize the average cost.
This requires taking the derivative of average cost. Then set the derivative equal to 0 and solve for x:

Answer 30

\[\overline{C(x)} = \frac{16}{\sqrt{x}} + \frac{x}{3375}\]
\[\large \overline{C(x)} = 16x^{-1/2}+ \frac{1}{3375}x\]
using power rule:
\[\large \overline{C'(x)} = 16\left( -\frac{1}{2}x^{-3/2}\right)+ \frac{1}{3375}\]
\[\large \overline{C'(x)} = -8x^{-3/2} + \frac{1}{3375}\]

Answer 31

so we set that equal to 0?

Answer 32

yep.

Answer 33

\[ \large -8x^{-3/2} + \frac{1}{3375} = 0\]
\[\large \frac{1}{3375} =8x^{-3/2}\]
multiply both sides by 1/8
\[\large \frac{1}{27000} =x^{-3/2}\]
then raise both sides to the -2/3 power
\[\large \left(\frac{1}{27000}\right )^{-2/3} =\left(x^{-3/2}\right) ^{-2/3}\]

Answer 34

and then the x^(-3/2) cancels out?

Answer 35

yep, should get x = 900

Answer 36

yes!

Answer 37

how do i get minimal average cost?

Answer 38

e) the minimal average cost
Replace x with 900 in the average cost function:
\[\overline{C(900)} = \frac{16}{\sqrt{900}} + \frac{900}{3375} = $0.80\]

Answer 39

thanks!

Answer 40

can you help me with a few more questions

Answer 41

boyfriend just got home...gonna chill for now :) Good luck!

Answer 42

thanks