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zaphod
An open rectangular box is to be made wit a square base, and its capacity is to be 4000cm^3. find the length of the side of the base when the amount of material used to make the box is as small as possible... PLEASE HELP :)
@satellite73 @Callisto @Omniscience @amistre64
|dw:1349789270142:dw| \[b*b*h=4000\]
and theres some calculus involved to find minimum stuff eh
\[b^2h=4000\] \[D[b^2h=4000]\] \[D[b^2h]=D[4000]\] \[D[b^2]h+b^2D[h]=D[4000]\] \[2bb'h+b^2h'=0\] its been awhile since i tried figuring this one out, but hheres my first idea :)
D is just notation for "take the derivative of" with respect to some arbitrary independant variable; at the moment
another equation we need to consider it "amount of material"
The amount of the material can be expressed through total surface area. Thus \[A_{t}=4bh+b^{2}\]But \[h=\frac{ 4000 }{ b^{2} }\] Now insert the expression of h in At, the take derivative.
using the surface area formula pforvided by Zek \[A=4bh+b^2\] \[D[A=4bh+b^2]\] \[D[A]=D[4bh]+D[b^2]\] \[A'=4(D[bh])+2bb'\] \[A'=4(D[b]h+bD[h])+2bb'\] \[A'=4(b'h+bh')+2bb'\] \[A'=4b'h+4bh'+2bb'\] this gives us 2 equations that work together \[A'=4b'h+4bh'+2bb'\]\[0=2bb'h+b^2h'\] just trying to recall if i needed to go thru all that ... if not just for the practice
It is not what I am saying. What I am trying too explain is we have the volume 4000 = hbb and At = 4hb+bb. Now combining this two equations we get\[A_{t}=\frac{ 1600 }{ b }+b^{2}\]Therfore take the derivative at this stage. Will u do that?
yes dA/db = 0 \[1600b^{-1} + b^{2}\] \[0 = -1600b^{-2} + 2b\] solve for b
That is what I am saying. Thanks
b = 20 thanks everyone:)
we can prolly assume we can work this by adjusting the base with respect to itself so b'=1 \[A'=4h+4bh'+2b\]\[0=2bh+b^2h'\] \[2b=\frac{-b^2h'}{h}\] \[A'=4h-2\frac{b^2h'}{h}h'-\frac{b^2h'}{h}\]using h=4000b^-2, h'=-8000b^-3 \[A'=4\frac{4000}{b^2}-2\frac{b^2\frac{-8000^2}{b^6}}{\frac{4000}{b^2}}-\frac{b^2\frac{-8000}{b^3}}{\frac{4000}{b^2}}\] \[A'=\frac{16000}{b^2}-\frac{32000}{b^2}+2b\] \[A'=2b-\frac{16000}{b^2}\] \[A'=\frac{2b^3-16000}{b^2}=0\] \[b=2(1000)^{1/3}=20\]
im sure that was not the simplest route to take lol