tmagloire1
  • tmagloire1
If 1,200 cm2 of material is available to make a box with a square base and an open top, find the maximum volume of the box in cubic centimeters. Answer to the nearest cubic centimeter without commas. For example, if the answer is 2,000 write 2000.
Mathematics
jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
is this a calculus problem?
tmagloire1
  • tmagloire1
yes calc ab.
tmagloire1
  • tmagloire1
I got 45,000

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amistre64
  • amistre64
what is your equation for volume?
tmagloire1
  • tmagloire1
300x - x^3/2
amistre64
  • amistre64
1,200 = b^2 + 4bh hmm, have you done langrange multipliers?
tmagloire1
  • tmagloire1
langrange multipliers? Never heard of them. this is just applied min and max problems
amistre64
  • amistre64
V = b^2h ; A = b^2+4bh-1200 = 0 Vb = 2bh; L~Ab = L(2b+4h) Vh = b^2; L~Ah = L(4b) equating partials we have 4Lb = b^2, b=4L 2bh = 2Lb+4Lh bh-2Lh = Lb h(b-2L) = Lb h = Lb/(b-2L) = 2L ---------------- 3L^2 = 75 L=5 -------------- b=20, h=10 is what im getting ... if i did a proper langrange :)
amistre64
  • amistre64
i tried doing a min max setup but i wasnt getting anywhere useful, so i attempted the langrange
tmagloire1
  • tmagloire1
Okay that's long-range that's what we do as well. so because b=20 and h=10 the volume then we get lwh=20(10)(5)=1000?
amistre64
  • amistre64
we could take our volume in terms of b given that b^2+4bh-1200 = 0 h = (1200-b^2)(4b)^(-1) V = b^2(1200-b^2)(4b)^(-1) or V = (1200b -b^3)/4
amistre64
  • amistre64
no, 20*20*10
tmagloire1
  • tmagloire1
Did you get the second 20 from b^2+4bh-1200 = 0
amistre64
  • amistre64
V = (1200b -b^3)/4 V' = (1200-3b^2)/4 = 0 1200-3b^2 = 0 1200 = 3b^2 400 = b^2 ; b=20
tmagloire1
  • tmagloire1
ohh ok i forgot to take the derivative of v.
amistre64
  • amistre64
no, i got b (the base) from the workings ... the base has a side length of 20
tmagloire1
  • tmagloire1
Okay, I understand now, thank you for helping me!
amistre64
  • amistre64
youre welcome

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