Show that the height of a closed right circular cylinder of maximum volume that can be inscribed
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hey mate
in two dimensions it looks like a rectangle inscribed in an isosceles triangle.
let the base of this triangle lie on the x-axis, with the two sides of the triangle symmetrical to the y-axis.
let the base of the triangle = 2R, height = H
side of the triangle in quad I has equation y = H - (H/R)x
base of the cylinder = x
height of the cylinder = y = H - (H/R)x
so, cylinder volume is ...
V =*x*[H - (H/R)x] = H[x- x/R]
dV/dx = H[2x - 3x/R]
set dV/dx = 0, x(2 - 3x/R) = 0
x = 2R/3 ... the cylinder radius for max volume
so height of the cylinder=y=H-(H/R)(2R/3)
y=H/3
plzz mark as brainliest answer
in two dimensions it looks like a rectangle inscribed in an isosceles triangle.
let the base of this triangle lie on the x-axis, with the two sides of the triangle symmetrical to the y-axis.
let the base of the triangle = 2R, height = H
side of the triangle in quad I has equation y = H - (H/R)x
base of the cylinder = x
height of the cylinder = y = H - (H/R)x
so, cylinder volume is ...
V =*x*[H - (H/R)x] = H[x- x/R]
dV/dx = H[2x - 3x/R]
set dV/dx = 0, x(2 - 3x/R) = 0
x = 2R/3 ... the cylinder radius for max volume
so height of the cylinder=y=H-(H/R)(2R/3)
y=H/3
plzz mark as brainliest answer
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