# 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?

Aug 6, 2016

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

#### Explanation:

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

$V = 4000 c f t = \left(\frac{2}{3} \pi {r}^{3} + \pi {r}^{2} h\right) = \pi {r}^{2} \left(\frac{2}{3} r + h\right) c f t$, So,

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

Cost $C = 2 \pi r h$ (for cylindrical wall surface)

$+ 5 \left(2 \pi {r}^{2}\right)$( for hemispherical dome surface) uniits

$= 2 \pi r \left(h + 5 r\right)$ units

Eliminating h,

$C \left(r\right) = 2 \pi r \left(\left(\frac{4000}{\pi {r}^{2}} - \frac{2}{3} r\right) + 5 r\right)$

$= 2 \left(\frac{4000}{r} + \frac{13}{3} \pi {r}^{2}\right)$ units.

$C ' \left(r\right) = 2 \left(- \frac{4000}{r} ^ 2 + \frac{26}{3} r\right)$ unitb/square foot.

For minimum C, C' = 0. So,

$- \frac{2000}{r} ^ 2 + \frac{13}{3} \pi r = 0$

$r = {\left(\frac{6000}{13 \pi}\right)}^{\frac{1}{3}} f e e t .$

$= 5.28 '$, nearly, and correspondingly,

$h = \cdot \left(\frac{4000}{\pi {r}^{2}} - \frac{2}{3} r\right)$, when r = 5.28'

=.42.15'.