# Lost on word problem

1. Mar 21, 2007

### willywonka12345

1. The problem statement, all variables and given/known data

The cost associated with storing x units of a certain product is given by the function C(x) = 0.8x^2 - 12.8x + 2600 where C is measured in dollars and x in hundreds of units. What is the minimum cost ?

3. The attempt at a solution

Umm, yea. I would attempt it but I have no idea where to start, but I will try anyways. So I was thinking it needs to be set up ( y = mx + b ) so I did and got a final result of : y = square root of 16x - 3250 OR : y = (16x - 3250)^.5

I then plotted it on a graph and figured that I need to find minimum x for some reason because y didn't make much sense as it came to be 0. So then the minimum x was about 204 on the graph when y = 0. So I was thinking that is my answer but have absolutely NO clue !! :rofl: Please help me !!

2. Mar 21, 2007

### anantchowdhary

Do you have any knowledge of derivatives.Please tell me if i can help....it wud be great if i could help ...i myslef have only limited knowledge.However.....$$C(x) = 0.8x^2 - 12.8x + 2600$$ is the equation we have.Now if we take $$y=c(x)$$

and differentiate it with respect to $$x$$...i think you would get ur answer setting $$\frac {dy}{dx}=0$$

3. Mar 21, 2007

### willywonka12345

No clue, we haven't got that far into the book. So that would tell me that it wouldn't be necessary to use that. But as before, still not sure. I wish you could help, I have a couple days to get the answer before returning to school so hopefully someone will know.

4. Mar 21, 2007

### anantchowdhary

Hm....i gues ud have to graph the equation y=C(x)= 0.8x^2 - 12.8x + 2600 !

5. Mar 21, 2007

### willywonka12345

And then what. lol

6. Mar 21, 2007

### anantchowdhary

find the curves minima point.I mean where the slope is 0!.

7. Mar 21, 2007

### willywonka12345

Which would be 2600. Surely that's not right.

8. Mar 21, 2007

### hage567

No, the slope is not zero at when C(x)=2600! Look harder.

9. Mar 21, 2007

### willywonka12345

and that's one of my weaknesses. So after looking at it again harder, I am getting something like x is around 8 and y is around 2548 but still am not sure as to what my answer is. 2548 dollars ?

Last edited: Mar 21, 2007
10. Mar 21, 2007

### Dick

If you want to do this without using derivatives and without using a graph, complete the square in the expression for C(x). Ie write it in the form:

C(x)=0.8*(x-A)^2+B

(Where you have to figure out what A and B are). Then clearly the minimum is where x=A and it's value is?

11. Mar 21, 2007

### willywonka12345

Ahhhhhhhhh, that is what I am talking about. Didn't think of the standard form dealio. Thanks Dick.

So C(x) = 0.8(x-6.4)^2 + 2559.04 , so the minimum you are saying would be 6.4 or 2559.04. I am thinking you are saying it is 6.4 because I believe you take the opposite sign of A in the (). Correct ? 6.4 ?

12. Mar 21, 2007

### Dick

The value of x at which the minimum occurs is x=6.4. At x=6.4 the cost C(x) reaches a minimum value of 2559.04. The question asks for minimum cost. (I'm trusting you got the numbers right - I didn't check them).

13. Mar 21, 2007

### Dick

No. You didn't get the numbers right. Multiply your C(x) back out. Do you get back to the original???

14. Mar 21, 2007

### willywonka12345

I didn't divide the .8 into 12.8. This one comes out right :

C(x) = 0.8(x-8)^2 + 2536 and then you kinda lost me as to finding the minimum cost.

15. Mar 21, 2007

### Dick

The squared term is always greater than or equal to zero. So the x value where it is zero defines the minimum of C(x). C(8)=2536 is the minimum cost. That's pretty close to what you were seeing on your graph.

16. Mar 21, 2007

### willywonka12345

Excellent, thanks so much for the help Dick. I like your way a whole lot better than the graph way !!

17. Mar 21, 2007

### Dick

Glad you like it. Wait till you get to derivatives. You'll LOVE them.

18. Mar 22, 2007

### saplingg

I lol'ed.

@OP:
Your textbook may have it listed as "completing the square", in case you were interested.