Question

A water tower is formed by joining two hemispheres to a cylinder and removing the flat surfaces at the ends. The volume needs to be 200 cubic meters. The metal used to form the hemispheres costs three times as much as the metal used to form the cylinder part. What dimensions of the water tower will minimize the cost?

Answer 1

Wait what?
removing the flat surfaces...
... oh!
I get it

okey dokey
basicaly find the shape that gives you a volume of 200 cubic meters and and minimizes the amount of materials needed to make the conainer


find the volume
we can take the 2 hemispheres and combine them to get 1 sphere
we can take the cylinder as well
vsphere=(4/3)pir^3
vcylinder=hpir^2
so total volume would be (4/3)pir^3+hpir^2


now find surface area which is the amount of sheet metal we will need to construct it
surface area=lateral area of cylinder+surface area of sphere
lateral area of cylinder=2pirh
surface aea of sphere=4pir^2 
total surface area=2pirh+4pir^2
but, the hemishpere part costs 3 times more
so just say the cylinder part costs 1 dollar per square meter and hemisphere part costs 3 dollars so we mutiply hemishpere part by 3 to get
SAcost=2pirh+12pir^2



so we got
v=200=hpir^2+4/3pir^3
solve for h since the r exponents are tricky
$h=\frac{200-\frac{4}{3} \pi r^3}{\pi r^2}$
subsitute that for h in the other equation
SAcost=2pirh+12pir^2
$SAcost=2 \pi r (\frac{200-\frac{4}{3}\pi r^3}{\pi r^2})+12 \pi r^2$
now we simpilify and take the derivitive to find the value of r where SAcost is a minimum
we get at that the derivitive of SAcost is 0 at r=$\sqrt[3]{\frac{150}{7 \pi}}$
by subsituteion we find that the value of h will be then $h=\frac{4(7 \pi -1)\sqrt[3]{7350}}{21\sqrt[3]{\pi}}$

aproximately
the radius should be about 1.8964m and the height should be about 53.0789m