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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    i think you should use inequalities, and finaly draw a graph using them. Right sstarica?

  2. anonymous
    • 5 years ago
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    hmmm, it's more like an optimization problem where you have to find the Absolute maximum in this case.

  3. anonymous
    • 5 years ago
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    he base is a square, but the other parts take shape of a cylinder, weird ._.

  4. anonymous
    • 5 years ago
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    the*

  5. anonymous
    • 5 years ago
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    I can't figure out a sketch for this one >_<, with no sketch, no formula, no answer sorry

  6. anonymous
    • 5 years ago
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    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..

  7. anonymous
    • 5 years ago
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    you mean L not I right?

  8. anonymous
    • 5 years ago
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    yeah, L..

  9. anonymous
    • 5 years ago
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    so, now we have the general function which is: V = w^2h

  10. anonymous
    • 5 years ago
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    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)

  11. anonymous
    • 5 years ago
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    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..

  12. radar
    • 5 years ago
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    Or an identity.

  13. radar
    • 5 years ago
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    The problem states only the base is made from material that is 12 sq material.

  14. radar
    • 5 years ago
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    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!!!

  15. radar
    • 5 years ago
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    Can we safely assume that the base is a square? i.e. length = width.

  16. anonymous
    • 5 years ago
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    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.

  17. amistre64
    • 5 years ago
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    use JBWeld :)

  18. amistre64
    • 5 years ago
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    do we assume that the sheet metal is a square to begin with? or simply 12m^2 of rectangular shape?

  19. amistre64
    • 5 years ago
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    most of these problems have you cut corners and fold the sides..... but that is an assumption for this one....

  20. anonymous
    • 5 years ago
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    for me, the assumption is that the total surface area is 12 m^2. :)

  21. amistre64
    • 5 years ago
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    Cant we just buy a tank thats already large enough to use? :)

  22. anonymous
    • 5 years ago
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    lol., then you are the consumer, not the designer.

  23. radar
    • 5 years ago
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    I say we go to the ag store and buy the tank.

  24. radar
    • 5 years ago
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    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!

  25. anonymous
    • 5 years ago
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    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.

  26. amistre64
    • 5 years ago
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    whats the measurements of the sheet metal tho? is it: 1,12 2,6 3,4 or some decimalized monstrosity?

  27. amistre64
    • 5 years ago
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    like 4sqrt(3), 4sqrt(3) ?

  28. amistre64
    • 5 years ago
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    uhh..that shoulda been 2sqrt(3) s but you know....the keyboard hates me :)

  29. amistre64
    • 5 years ago
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    also, it just says tha the base is square; it doesnt say that the tank itself is square.... hmmmm

  30. anonymous
    • 5 years ago
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    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. ^_^

  31. radar
    • 5 years ago
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    That works out for the amount of material.....but is it the maximum volume, I'm thinking it is but can't prove it.

  32. anonymous
    • 5 years ago
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    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

  33. radar
    • 5 years ago
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    That would be w=2 I guess that is the proof that is needed. Thanks for the step by step procedure.

  34. radar
    • 5 years ago
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    \[V=(12w-w ^{3})/4=3w-w ^{3}/4=3w-1w ^{3}/4\]

  35. radar
    • 5 years ago
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    \[dV/dw=3-3w ^{2}/4\]

  36. anonymous
    • 5 years ago
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    wow., how did you do that radar?

  37. radar
    • 5 years ago
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    Set to 0 \[3-3w ^{2}/4=0\] \[3w ^{2}/4=3\] Multiplying both sides by 4/3 \[w ^{2}=4\]

  38. radar
    • 5 years ago
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    Hey, Janjor did the ground work! lol

  39. anonymous
    • 5 years ago
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    yeah! i think i don't understand what you mean by "ground" lol

  40. radar
    • 5 years ago
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    Set up the foundation (foundations are usually on the ground)

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