Posted by **Megan** on Friday, July 8, 2011 at 11:41am.

Let C(x) = 0.02x^3 + 55x^2 + 1380.

Find the minimum average cost for this commodity.

I got to the point where I got x=-1374.98, x=-5.018, x=5

but what do I have to do again after this? Because I am missing some steps after I find x.

Do i have to plug all the X's to the original?

thanks :)

- You have a typo -
**Damon**, Friday, July 8, 2011 at 3:41pm
You have a typo. You need a - sign somewhere on the right.

dC/dx = 0 at min or max

.02(3)x^2 + 55(2) x = 0

x (.06 x + 110 ) = 0

x = 0 or x = - 1833

0 or a negative number does not makes sense

Your cost has no minimum in the first quadrant, it goes up forever with number of units.

- Math Calculus -
**Megan**, Friday, July 8, 2011 at 4:08pm
no i don't see any negative sign in my homework question...

first I did C*bar(x)/x = 0.02x^3+55x^2+1380/(x)

= 0.02x^2 + 55 x + 1380/x

then I differentiate it C*bar*'(x) = 0.4x - 1380/x^2 +55

then I set it equal to zero which gives me x = -137.317 or x=-5.104 and 4.921

I did the solve function with the calculator

- Megan ! - Math Calculus -
**Reiny**, Friday, July 8, 2011 at 8:57pm
Megan, if you look at your original post, there was no division by x at the end

So Damon was right to question your typing

so

C(x) = (0.02x^3 + 55x^2 + 1380)/x

= .02x^2 + 55x + 1380/x

C'(x) = .04x + 55 - 1380/x^2

= 0 for a max/min of C

.04x^3 + 55x^2 - 1380 = 0

solving this with my online equation solver I got

x = 5 and two negative answers

so x = 5

plug that back into original C(x)

- Math Calculus -
**Megan**, Saturday, July 9, 2011 at 11:05am
Reiny,

Before you do equals to zero, how did you get 0.04x^3 + 55x^2 - 1380 = 0?

I set

0.04x + 55 - 1380/x^2 = 0

and I got x=-137.317, or x=-5.204,, or x=4.921

if i plus in to the originals i have to plus each x's?

and the minimum average is the smallest number from the result after I plug in the Xs?

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