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