Cost of manufacturing problem

  • #1
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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?
 

Answers and Replies

  • #2
1,235
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Ok so [tex] C(x) = 625+15x+0.01x^2 [/tex] for [tex] 1 \leq x \leq 500 [/tex] The unit cost is [tex] \frac{625}{x} + 15+0.01x [/tex]. So find derivative of [tex] \frac{C(x)}{x} [/tex] and set it equal to 0 to find critical points. And then find your minimum
 
  • #3
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k i've got 0= -0.01+(15)x^-1+(625)x^-2

NOW WHAT....how to find x
 
  • #4
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[tex] \frac{dU}{dx} = \frac{-625}{x^2} + 0.01 [/tex]. Now solve for x.
 
  • #5
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dude what happened to the 15x^-1 !!!!!
 
  • #6
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when you have a constant thomas such as 15, you dont consider it as [tex]\frac{15}{x^0}[/tex] and differentiate as usual. Constants simply disappear when differentiated with respect to a variable. So you dont have [tex]\frac{15}{x}[/tex].

One thing you have to be careful about; when you set [tex]\frac{dU}{dx}=0[/tex] 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
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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
HallsofIvy
Science Advisor
Homework Helper
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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
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k thanks i got it
 

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