Tough Question

[thomasrules]

Suppose that the cost of manufacturing x items is approximated by C(x)=625 + 15x +0.01x^2, for 1 < or equal to x < or equal to 500. The unit cost would then be U(x) = C(x)/x. How many items should be manufactured in order to ensure that the unit cost is minimized.....

I DONT KNOW WHAT TO DO.....

I started off by doing the derivative of the first equation, which was a guess but then what I do?

[courtrigrad]

Ok so C(x) = 625+15x+0.01x^2 for 1 \leq x \leq 500 The unit cost is \frac{625}{x} + 15+0.01x . So find derivative of \frac{C(x)}{x} and set it equal to 0 to find critical points. And then find your minimum

[thomasrules]

k i've got 0= -0.01+(15)x^-1+(625)x^-2

NOW WHAT....how to find x

[courtrigrad]

\frac{dU}{dx} = \frac{-625}{x^2} + 0.01 . Now solve for x.

[thomasrules]

dude what happened to the 15x^-1 !!!!!

[Ryoukomaru]

when you have a constant thomas such as 15, you dont consider it as \frac{15}{x^0} and differentiate as usual. Constants simply disappear when differentiated with respect to a variable. So you dont have \frac{15}{x}.

One thing you have to be careful about; when you set \frac{dU}{dx}=0 you are looking for maximuns and minimums, you might also need to do a 2nd Derivative test to find out which one it is.

[thomasrules]

ok so if my equation was right: 0=-0.01+(15)x^-1+(625)x^-2

Then I did what you suggested ryoukomaru and did the second derivative of that.

I got 0= -15x^-2-1250x^-3

I HATE THIS PLEASE HELP

[HallsofIvy]

No, your equation is NOT right- that's what Ryoukumaru was telling you. He said "you DON'T have 15/x"!

C(x)/x= 625/x+ 15+ 0.01x2

The derivative of 625/x= 625x-1= -625x-2.
The derivative of 15, a constant, is 0!
The derivative of 0.01x2 is 0.02 x.

The derivative of C(x)/x= -625-2+ 0.02x. Set that equal to 0 and solve for x.

[thomasrules]

k thanks i got it