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monet A
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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.
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