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'.