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Homework Help: Cost of manufacturing problem

  1. Jan 26, 2005 #1
    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 started off by doing the derivative of the first equation, which was a guess but then what I do?
  2. jcsd
  3. Jan 26, 2005 #2
    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
  4. Jan 26, 2005 #3
    k i've got 0= -0.01+(15)x^-1+(625)x^-2

    NOW WHAT....how to find x
  5. Jan 26, 2005 #4
    [tex] \frac{dU}{dx} = \frac{-625}{x^2} + 0.01 [/tex]. Now solve for x.
  6. Jan 26, 2005 #5
    dude what happened to the 15x^-1 !!!!!
  7. Jan 26, 2005 #6
    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.
  8. Jan 26, 2005 #7
    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

  9. Jan 26, 2005 #8


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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.
  10. Jan 26, 2005 #9
    k thanks i got it
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