# minimization of a water tank

• Jan 19th 2008, 09:47 AM
winterwyrm
minimization of a water tank
Thanks in advance, it's attached. In case you can't read it, the problem states: The tank below is to be constructed to hold 5.00 x 10^5L when filled completely. The shape is cylindrical with a hemispherical top. The cost is a function of surface area, it will cost $300/(m^2) for the cylindrical portion, and$400/(m^2) for the hemispherical portion. Ignore the cost to construct the bottom circular area.

Specify the dimensions, H & R which will result in the lowest dollar cost, where H is the height of the cylinder, and R is the radius of the sphere and of the cylinder.
• Jan 19th 2008, 10:55 AM
galactus
The volume of the tank is:

$V=\frac{2}{3}{\pi}r^{3}+{\pi}r^{2}h=500,000$......[1]

The surface area of the hemisphere is given by $2{\pi}r^{2}$

The surface area of the cylidrical portion, excluding the base is given by:

$2{\pi}rh$

So, the total surface area is:

$S=2{\pi}r^{2}+2{\pi}rh$

But, the cost for the hemsphere is $400 and the cylinder is$300:

$S=400(2{\pi}r^{2})+300(2{\pi}rh)$....[2]

Solve [1] for, say, h and sub into [2]. Then differentiate, set to 0 and solve for r. h will follow.
• Jan 19th 2008, 03:25 PM
winterwyrm
when I do it this way though, I get R= 457.08m and H= 913.08m though, and those just don't work in the original equation, is this right?