## Tough Question

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?
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 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
 k i've got 0= -0.01+(15)x^-1+(625)x^-2 NOW WHAT....how to find x

## Tough Question

$$\frac{dU}{dx} = \frac{-625}{x^2} + 0.01$$. Now solve for x.
 dude what happened to the 15x^-1 !!!!!
 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.
 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
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.
 k thanks i got it