anonymous
  • anonymous
An open metal tank with a square base is made from 12m^2 of sheet metal. Find the length of the side of the base for the volume of the tank to be a maximum and find this maximum volume
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i think you should use inequalities, and finaly draw a graph using them. Right sstarica?
anonymous
  • anonymous
hmmm, it's more like an optimization problem where you have to find the Absolute maximum in this case.
anonymous
  • anonymous
he base is a square, but the other parts take shape of a cylinder, weird ._.

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More answers

anonymous
  • anonymous
the*
anonymous
  • anonymous
I can't figure out a sketch for this one >_<, with no sketch, no formula, no answer sorry
anonymous
  • anonymous
maxima problem.. determine what subject to be maximize or minimize; in this case, the volume of a tank.. V = lwh and in this case the l is equal to the w. therefore V = w^2*h then, give the constraint of the problem.. in this case; the material to be use.. only 12 m^2 Make an equation for the constraint: that is, the formula for the surface area..
anonymous
  • anonymous
you mean L not I right?
anonymous
  • anonymous
yeah, L..
anonymous
  • anonymous
so, now we have the general function which is: V = w^2h
anonymous
  • anonymous
the constraint: the surface area of a tank, cube with open on the top, which means there are 5 faces only. one on the bottom, and four on the sides. The surface area on the bottom is w^2 and the surface area on the side is wh. since there are 4 side faces, 4w^2 therefore the total surface area is: w^2 + 4wh = 12 m^2(meter squared)
anonymous
  • anonymous
from the general function stated by sstarica, express h in terms of w from my total surface area equation, then we will have a function of V in terms of w..
radar
  • radar
Or an identity.
radar
  • radar
The problem states only the base is made from material that is 12 sq material.
radar
  • radar
Never mind it does probably include the sides in the that 12 sq meters of material. Come on fellows go ahead and work this out. I can't envision construction of this tank without some waste in the material (cutting out the corners). I guess they could just press it into a mold!!!
radar
  • radar
Can we safely assume that the base is a square? i.e. length = width.
anonymous
  • anonymous
as what i understand the problem. the tank is square base, and the tank is made from 12m^2. The problem does not define how the tank is constructed.
amistre64
  • amistre64
use JBWeld :)
amistre64
  • amistre64
do we assume that the sheet metal is a square to begin with? or simply 12m^2 of rectangular shape?
amistre64
  • amistre64
most of these problems have you cut corners and fold the sides..... but that is an assumption for this one....
anonymous
  • anonymous
for me, the assumption is that the total surface area is 12 m^2. :)
amistre64
  • amistre64
Cant we just buy a tank thats already large enough to use? :)
anonymous
  • anonymous
lol., then you are the consumer, not the designer.
radar
  • radar
I say we go to the ag store and buy the tank.
radar
  • radar
If we assume only the base is 12 sq meters then we have no limit on the height of the sides that means you could have a mighty big tank!
anonymous
  • anonymous
yeah, that is why it must be assumed that the total surface area of the tank is 12 m^2. I think I already give the equation for that.. using that assumption we can have a solution.
amistre64
  • amistre64
whats the measurements of the sheet metal tho? is it: 1,12 2,6 3,4 or some decimalized monstrosity?
amistre64
  • amistre64
like 4sqrt(3), 4sqrt(3) ?
amistre64
  • amistre64
uhh..that shoulda been 2sqrt(3) s but you know....the keyboard hates me :)
amistre64
  • amistre64
also, it just says tha the base is square; it doesnt say that the tank itself is square.... hmmmm
anonymous
  • anonymous
I tried to solved it. the answer is: width = 2m length = 2 m (for square base) height = 1 m therefore the total volume is 4 m^3, the total surface area is 12 m^2. ^_^
radar
  • radar
That works out for the amount of material.....but is it the maximum volume, I'm thinking it is but can't prove it.
anonymous
  • anonymous
try use the equation: V = w^2 * h (square base) Surface area = 12 m^2 = w^2 (square base) + 4 * w * h (four faces of the sides of the tank) solve for h in terms of w: h = (12 - w^2) / 4w therefore, the volume becomes V = (w^2) * (12-w^2)/4w or V = (12w - w^3)/4 then derive V with respect to w
radar
  • radar
That would be w=2 I guess that is the proof that is needed. Thanks for the step by step procedure.
radar
  • radar
\[V=(12w-w ^{3})/4=3w-w ^{3}/4=3w-1w ^{3}/4\]
radar
  • radar
\[dV/dw=3-3w ^{2}/4\]
anonymous
  • anonymous
wow., how did you do that radar?
radar
  • radar
Set to 0 \[3-3w ^{2}/4=0\] \[3w ^{2}/4=3\] Multiplying both sides by 4/3 \[w ^{2}=4\]
radar
  • radar
Hey, Janjor did the ground work! lol
anonymous
  • anonymous
yeah! i think i don't understand what you mean by "ground" lol
radar
  • radar
Set up the foundation (foundations are usually on the ground)

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