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

- anonymous

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- anonymous

i think you should use inequalities, and finaly draw a graph using them. Right sstarica?

- anonymous

hmmm, it's more like an optimization problem where you have to find the Absolute maximum in this case.

- anonymous

he base is a square, but the other parts take shape of a cylinder, weird ._.

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## More answers

- anonymous

the*

- anonymous

I can't figure out a sketch for this one >_<, with no sketch, no formula, no answer
sorry

- 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

you mean L not I right?

- anonymous

yeah, L..

- anonymous

so, now we have the general function which is:
V = w^2h

- 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

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

Or an identity.

- radar

The problem states only the base is made from material that is 12 sq material.

- 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

Can we safely assume that the base is a square? i.e. length = width.

- 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

use JBWeld :)

- amistre64

do we assume that the sheet metal is a square to begin with? or simply 12m^2 of rectangular shape?

- amistre64

most of these problems have you cut corners and fold the sides..... but that is an assumption for this one....

- anonymous

for me, the assumption is that the total surface area is 12 m^2. :)

- amistre64

Cant we just buy a tank thats already large enough to use? :)

- anonymous

lol., then you are the consumer, not the designer.

- radar

I say we go to the ag store and buy the tank.

- 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

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

whats the measurements of the sheet metal tho? is it:
1,12
2,6
3,4
or some decimalized monstrosity?

- amistre64

like 4sqrt(3), 4sqrt(3) ?

- amistre64

uhh..that shoulda been 2sqrt(3) s but you know....the keyboard hates me :)

- amistre64

also, it just says tha the base is square; it doesnt say that the tank itself is square.... hmmmm

- 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

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

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

That would be w=2 I guess that is the proof that is needed. Thanks for the step by step procedure.

- radar

\[V=(12w-w ^{3})/4=3w-w ^{3}/4=3w-1w ^{3}/4\]

- radar

\[dV/dw=3-3w ^{2}/4\]

- anonymous

wow., how did you do that radar?

- radar

Set to 0
\[3-3w ^{2}/4=0\]
\[3w ^{2}/4=3\]
Multiplying both sides by 4/3
\[w ^{2}=4\]

- radar

Hey, Janjor did the ground work! lol

- anonymous

yeah!
i think i don't understand what you mean by "ground"
lol

- radar

Set up the foundation (foundations are usually on the ground)

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