Question

For the given cost function "C(x)=19600+300x+x^2" find:

a) The production level that will minimize the average cost
b) The minimal average cost

Any help is appreciated. TIA

Answer 1

u knw hw to differentiate the function?

Answer 2

Yes I do. C'(x) = 2(x+150) I believe

Answer 3

\[C'(x)=300 + 2x\]

Answer 4

it is indeed. on differentiating it for a second time, u get just 2, which is greater than zero which holds true for minima. so equate the first derivative to zero & find x
\[2(x+150)=0\]

Answer 5

x = -150 in that case

Answer 6

which makes x = -150. put that in C(x) function. that will give u the min cost. as for production level, m not sure about that to be honest.

Answer 7

Ah, no worries, thank you for your help!

Answer 8

welcome... but did u understand the production level question anyway?

Answer 9

Actually, I just tried it, and it appears to be incorrect. I plugged "-150" into the C(x) function to get "-2900".

Answer 10

yes. seems theres something wrong. hold on!

Answer 11

okay, m pretty sure that the method followed is correct. minimizing the function does give -150. ur question seems to have some fault. re check it

Answer 12

I've looked over my question and it is worded exactly as in the original post :/

Answer 13

then the question itself is wrong. i just checked with wolframalpha. computers cant be wrong :/
http://www.wolframalpha.com/input/?i=minimize+19600+%2B+300x+%2Bx%5E2

Answer 14

I suspect it has something to do with the average cost? Because right now, we're only deriving the cost function and not working with the average cost...

Answer 15

When you are minimizing average cost, you need to divide the cost function by x then differentiates and set it equal to zero to fine the minimal

Answer 16

now if u'd only mind explaining what u just said? no pun intended.

Answer 17

"19600+300x+x^2" / "x" = "(19600/x)+300+x"

Derivative of: "(19600/x)+300+x" is "1-(19600/x^2)"

1-(19600/x^2)?

Answer 18

its by definition of average you add up all the numbers and divide by the amount

you are only given the cost function not the average cost function

Answer 19

yea that is the correct derivative

Answer 20

part a is the optimal x value
and part b is the optimal y value

Answer 21

solutions will be -1/140 & +1/140

Answer 22

And to get the desired result of 0, we just find the square-root of 19600, so that the 1 can subtract into 1 to get 0? So the answer should be "140"?

Answer 23

& on finding the second derivative:
\[C''(x)=-39200x\]

Answer 24

so the value of x that will be used in the original function(the one that has been divided by x) is -1/140

Answer 25

the answer i got is 140 for the optimal x value

Answer 26

it doesnt really matter in this case since there is a 'square' but u hav to keep that in mind

Answer 27

no wait, i screwed up again :P

Answer 28

140 is indeed the value to be used! @jayz657

Answer 29

While we know that 140 is the optimal x value, I'm uncertain as to how we can find out the minimal average cost.

Answer 30

\[C''(x)=39200/x^3\]

Answer 31

the minimal average cost is when you just plug in 140 for the x value

Answer 32

Do we just plug 140 into the cost function?

Answer 33

yea thats right

Answer 34

yes

Answer 35

answer is 0 btw :3

Answer 36

you need to do C(140)

Answer 37

Average cost function that is. The answer is 580 I believe.

Answer 38

your answer should be bigger than 19,200

Answer 39

19600 sorry

Answer 40

so u plugged 140 in \[19600 + 300x + x^2 \]
or
\[19600/x +300 +x\]

Answer 41

crap sorry i messed up your right its 580

Answer 42

its the second equation

Answer 43

We are looking for: b) The minimal average cost

So I plugged "140" into the average cost function and got "580"

Answer 44

Ah, glad we're on the right page. Thank you for your help today folks, it's appreciated. :)

Answer 45

your welcome glad to help

Answer 46

@jayz657 thanks from my side too ^_^ !