Bunuel wrote:
What is the volume of a right cylindrical can X?
(1) If the height of can X were equal to the radius of the base of can X, the volume of the new cylinder would have been half of the volume of can X
(2) If the radius of the base of can X were equal to height of can X, the volume of the new cylinder would have been four times the volume of can X
Volume of right cylindrical can X = pi rx^2 hx
Vn new volume
Vi initial volume
rx and hx are radius and height of the cylinder, respectively
(1) If the height of can X were equal to the radius of the base of can X, the volume of the new cylinder would have been half of the volume of can X
Now if hx = rx, then
Vn = 1/2 Vi
2 Vn = Vi
2 pi rx^3 = pi rx^2 *hx
2 pi rx^3 - pi rx^2 *hx = 0
rx^2 ( 2rx - hx) = 0
2rx = hx, cant say anything from here
(2) If the radius of the base of can X were equal to height of can X, the volume of the new cylinder would have been four times the volume of can X
Now if rx = hx
Vn = 4 Vi
pi rx^3 = 4 pi rx^2 *hx
4 pi rx^2 *hx - pi rx^3 = 0
rx^2 ( 4hx - rx) = 0
4hx = rx, cant say anything from here
From combination of both, i can
4hx = rx and 2rx = hx
Will we get a definite value from this, No i doubt it .
E
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