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anonymous

  • 5 years ago

An open box is being constructed whose base length is 3 times the base width and whose volume is 50 cubic meters. If thematerials used to build the box cost $10 per square meter for the bottom and $6 per square meter for the sides, what are the dimensions for the least expensive box? Help?!?! I feel as if not enough information is provided. Yet again I could be wrong? How do I solve this problem?

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  1. anonymous
    • 5 years ago
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    ok ill help you :)

  2. anonymous
    • 5 years ago
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    this is an optimization problem, did you see that in class?

  3. anonymous
    • 5 years ago
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    Yes I do recognize that its an optimization problem, in which we have to alter our formulas.

  4. anonymous
    • 5 years ago
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    Help?

  5. anonymous
    • 5 years ago
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    do you need an answer?

  6. anonymous
    • 5 years ago
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    or set up

  7. anonymous
    • 5 years ago
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    setting up and understanding, please.

  8. anonymous
    • 5 years ago
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    okay, so first lets take care of the constraint. i.e. the volume of this box is 50 cm^3

  9. anonymous
    • 5 years ago
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    yes

  10. anonymous
    • 5 years ago
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    how about \[V=3x^2y\] and we know that \[50=3x^2y\]

  11. anonymous
    • 5 years ago
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    is that good?

  12. anonymous
    • 5 years ago
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    where did the 3x^2 come from?

  13. anonymous
    • 5 years ago
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    woah sorry i had to go fix a computer ill let ebbflo take over hes got a good flow going,

  14. anonymous
    • 5 years ago
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    ** I meant 3x^2y

  15. anonymous
    • 5 years ago
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    the length is 3 time the width

  16. anonymous
    • 5 years ago
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    so 3x^2 is the area of the base and I just let y be the heigth of the box

  17. anonymous
    • 5 years ago
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    is that good?

  18. anonymous
    • 5 years ago
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    yes

  19. anonymous
    • 5 years ago
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    okay now we need the surface area formula for this box which since there is no top is \[S=3x(x)+2(3x)y+2xy\]

  20. anonymous
    • 5 years ago
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    Area = base * height so the base =3x and the height=x

  21. anonymous
    • 5 years ago
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    i am just adding up the areas of the sides of the box

  22. anonymous
    • 5 years ago
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    the base of the box has dimensions (3x)cm by (x)cm

  23. anonymous
    • 5 years ago
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    are you good with the surface area formula I presented?

  24. anonymous
    • 5 years ago
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    If so We can proceed to the cost function

  25. anonymous
    • 5 years ago
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    im sorry meters

  26. anonymous
    • 5 years ago
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    can you explain to me how you arrived to your Surface area

  27. anonymous
    • 5 years ago
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    since this box only has 5 sides we just add up the areas of those sides the area of the bottom is (3x)(x) the combined area of one of the pairs of equal sides is 2(3x)(y) and the combined area of the remaining pair of equal sides is 2(x)(y) then I just summed those up

  28. anonymous
    • 5 years ago
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    so that's where I got my formula for the surface area

  29. anonymous
    • 5 years ago
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    allright.

  30. anonymous
    • 5 years ago
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    so now ti get the cost function we just multiply 10 by the area of the bottom and 6 times each of the combined areas of the sides like so \[C(x,y)=10(3x^2)+6(6xy)+6(3xy)\] collecting like terms we pretty it up to get \[C(x,y)=30x^2+48xy\]

  31. anonymous
    • 5 years ago
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    sorry I made a slight mistake in that formula

  32. anonymous
    • 5 years ago
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    the last one I wrote was correct

  33. anonymous
    • 5 years ago
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    \[C(x,y)=10(3x^2)+6(6xy)+6(2xy)\] should have been the first one from which I got \[C(x,y)=30x^2+48xy\]

  34. anonymous
    • 5 years ago
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    now we are ready to write the cost function above as a function of x only by way of the constraint

  35. anonymous
    • 5 years ago
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    okay.

  36. anonymous
    • 5 years ago
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    so since \[50=3x^2y\]\[y=\frac{50}{3x^2}\]

  37. anonymous
    • 5 years ago
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    so \[C(x)=30x^2+48x(\frac{50}{3x^2})\] and cleaning it up we get \[C(x)=30x^2+\frac{800}{x}\]

  38. anonymous
    • 5 years ago
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    if we are good, then now it is time to derive

  39. anonymous
    • 5 years ago
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    \[C^\prime(x)=60x-\frac{800}{x^2}\] get a common denominator to make finding the critical points easier we get\[C^\prime(x)=\frac{60x^3-800}{x^2}\]

  40. anonymous
    • 5 years ago
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    since this is a box with known volume we know \[x\neq 0\] so just solve \[60x^3-800=0\] for x

  41. anonymous
    • 5 years ago
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    we get \[x=2(\frac{5}{3})^\frac{1}{3}\]

  42. anonymous
    • 5 years ago
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    this is the only value we get so many people quit there and just go with but you can confirm this is a minimum value via the second derivative test for max/min

  43. anonymous
    • 5 years ago
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    use that vale to get the other dimensions and you are done...

  44. anonymous
    • 5 years ago
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    thank you soo much

  45. anonymous
    • 5 years ago
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    glad it helps, this is easier to explain with the aid of a sketch

  46. anonymous
    • 5 years ago
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    I was just looking over the work and you told me to use my x value to figure out my other dimensions. Do i do that by taking the second derivative?

  47. anonymous
    • 5 years ago
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    no just find the height by plugging that x value into the constraint we solved for why and the other dimension is just 3 time that x value we got

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