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Max/Min problem I'm fundamentally flawed in my understanding I think

  1. May 8, 2005 #1
    :cry:

    I am entirely frustrated with maximum and minimum problems. I have had this issue with them since I first was introduced to them and I thought I had resolved it but it keeps getting up and biting me again.

    The problem is I am not even sure what I am in the habit of doing wrong, so I can't fix it. Could somebody point out what I have done ar*eways here for me so I can get over this once and for all..?

    A closed box is to be made with length equal to 3 times its width. The total surface area of the box will be 30m^2. Find the dimensions that give maximum volume in the box.

    [tex] 3000cm^2 = 2lw + 2hw + 2lh [/tex]

    [tex] \frac {3000 - 6w^2} {8w}} = h [/tex]

    [tex] V = l *w * h [/tex]

    [tex] V(w) = 3w * w * \frac {3000 - 6w^2} {8w}} = \frac {9000w - 18w^3} {8}} [/tex]

    [tex] V'(w) = \frac {9000 - 54w^2} {8}}[/tex]

    [tex] V''(w) = \frac {- 108w} {8}} [/tex]

    Now when I find the 0 value of dv/dw it is one value that can be positive or negative, naturally I disregard the negative value, but then plugging anything positive into the second derivative I will always get a negative slope, so all values >0 would return as a maximum. I can finish off the problem from there and get values.

    However...

    My problem is that if I work it out in units of metres by allowing 30m^2 to equal the surface area I get a value of ca 1.29 metres, and if I use cm^2 (above) as the surface area unit I get ca 12.9cm. Both work to give 30m^2 as the surface area but they are clearly not the same values yet they are working perfectly in the same equation??????

    Should I have found both of these as critical points when looking for 0 in my 1st derivative? If so, how, what did I miss....

    please someone help me iron out this crinkle in my understanding because as you can see I am too confused to do it myself.
     
    Last edited: May 8, 2005
  2. jcsd
  3. May 8, 2005 #2

    Galileo

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    30 square meters is not equal to 3.000 square centimeters but 300.000 square centimeters.

    [tex]1 \mbox{m}^2 = 10.000 \mbox{cm}^2[/tex]

    you have to reduce both dimensions by a factor of 100.
     
  4. May 8, 2005 #3

    Oh :confused: as in 100 cm * 100 cm = 10000 cm^2 I thought it was a fundamental flaw in my understanding but *that* fundamental is just embarrassing, thanks for pointing it out to me :yuck: I think :blushing:
    I am sure I won't let this happen again, I might get paranoid enough to question all my chemistry calculations though... what have I done?
     
  5. May 8, 2005 #4
    don't worry, if you pay attention, you will see that people make this mistake all the time. pretty much any news article you read that converts units^2 does it incorrectly.

    to make things worse even, in my 300 level geography class a few weeks ago, the instructor was saying something about how many square kilometers of rainforest were destroyed each year. I think it was like, 25,000km^2 were destroyed. he's like "anyone got a calculator, so we can put that in miles?". and this nerdly kid calculates it, and comes up with 14,000mi^2 (which is wrong, and dumb to boot, since he had a ti-8x with the conversion built in..). to make things worse, the instructor says "so, that's about 7,000miles by 7,000 miles."

    yup, good thing i'm at a high quality university.
     
  6. May 8, 2005 #5

    quasar987

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    lol fishhy :biggrin:

    monet: A good trick not to get confused about unit conversion is to do this: 1m = 100cm, so

    [tex]30m^2 = 30 (100 cm)^2 = 30\cdot 100^2 \cdot cm^2 = 300000cm^2[/tex]

    cool huh?

    So as soon as you know the "linear conversion", you can retrieve any other "higher degree" conversion. Another example, taken from physics: 1eV = 1.6*10^-19 J, so

    [tex]5\cdot 10^9eV^3 = 5\cdot 10^9(1.6\cdot 10^{-19} J)^3 = 5\cdot 10^9\cdot (1.6\cdot 10^{-19})^3 J^3 = 8\cdot 10^{-10} J^3[/tex]
     
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