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12 Nov 2019, 02:04
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Difficulty:
65%
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Question Stats:
51% (01:37) correct
49%
(01:40)
wrong
based on 196
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A right cone is to be placed within a rectangular box so that the cone stands upright when the box is placed on one of its sides. If the dimensions of the box are 2 inches by 2 inches by 4 inches, then what is the greatest possible volume of such a cone?
A. \((\frac{2}{3})π\)
B. \((\frac{4}{3})π\)
C. \(2π\)
D. \((\frac{8}{3})π\)
E. \(4π\)
Are You Up For the Challenge: 700 Level Questions
A. \((\frac{2}{3})π\)
B. \((\frac{4}{3})π\)
C. \(2π\)
D. \((\frac{8}{3})π\)
E. \(4π\)
Are You Up For the Challenge: 700 Level Questions
Archit3110
Updated on: 12 Nov 2019, 13:37
Originally posted by Archit3110 on 12 Nov 2019, 03:26.
Last edited by Archit3110 on 12 Nov 2019, 13:37, edited 1 time in total.
Last edited by Archit3110 on 12 Nov 2019, 13:37, edited 1 time in total.
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vol of cone ; 1/3 * pi * r^2 *h
r has to be max to get max vol ; let side of base = 2 ; r = 1
h = 4
we get vol=
\((\frac{4}{3})π\)
IMO B
r has to be max to get max vol ; let side of base = 2 ; r = 1
h = 4
we get vol=
\((\frac{4}{3})π\)
IMO B
Bunuel
The maximum possible diameter of cone that can fit in the box, d= max{min(2,4), min(2,4), min(2,2)}=2
radius, r= 1 inches
height, h= 4 inches
Volume= \(\frac{1}{3}* pi* r^2*h\)= \(\frac{1}{3}*pi*1^2*4\)= \(\frac{4}{3} pi\)
radius, r= 1 inches
height, h= 4 inches
Volume= \(\frac{1}{3}* pi* r^2*h\)= \(\frac{1}{3}*pi*1^2*4\)= \(\frac{4}{3} pi\)
Bunuel
