# Cost of manufacturing problem

1. Jan 26, 2005

### 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?

2. Jan 26, 2005

### 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

3. Jan 26, 2005

### thomasrules

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

NOW WHAT....how to find x

4. Jan 26, 2005

### courtrigrad

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

5. Jan 26, 2005

### thomasrules

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

6. Jan 26, 2005

### 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.

7. Jan 26, 2005

### 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

8. Jan 26, 2005

### HallsofIvy

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.

9. Jan 26, 2005

### thomasrules

k thanks i got it

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