An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the hemispherical dome costs 5 times as much per square foot as the cylindrical wall, what are the most economic dimensions for a volume of 4000 cubic?

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Answer 1 · 2016-08-06T04:47:27

Radius oh dome = 5.28'= 5'3.4'', nearly, and the height of the cylinder = 42.15' = 42'2'', nearly

Let r' = the radius of the dome and h' = the height of the right circular

cylinder and cost per square foot surface area = 1 unit of cost. Then,

Volume enclosed

#V = 4000 cft = (2/3pir^3+pir^2h)= pir^2(2/3r+h) cft #, So,

#h =.(4000/(pir^2)-2/3r)'.

Cost #C = 2pirh# (for cylindrical wall surface)

#+5(2pir^2)#( for hemispherical dome surface) uniits

#=2pir(h+5r)# units

Eliminating h,

#C(r) = 2pir((4000/(pir^2)-2/3r)+5r)#

#= 2(4000/r+13/3pir^2)# units.

#C'(r) =2 (-4000/r^2+26/3r)# unitb/square foot.

For minimum C, C' = 0. So,

#-2000/r^2+13/3pir = 0#

#r = (6000/(13pi))^(1/3) feet.#

#=5.28'#, nearly, and correspondingly,

#h = *(4000/(pir^2)-2/3r)#, when r = 5.28'

#=.42.15'.