Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,561
Hi,Hi,
So they are asking to us optimize cost in terms of the surface area of the tank.
The equation for the surface of the tank is the surface of a sphere (2 hemispheres put together) and the surface area of the cylinder (ignoring the circle top and bottom);
SA=4Pir^2+2Pirh
Now to make the cost equation we just have to multiply the surface ara of the sphere by two, indicating the cost:
C = 8Pir^2+2Pirh
We can see we have 2 variables so we need to make a second equation. We are given the volume, 200m^3, so we can create the equation for the volume of the tank by adding the volume of the cylinder plus the volume of a sphere to give:
200 = 4/3Pir^3 + Pir^2h
Solve for h so we can input it in the cost equation, giving us only one variable;
h = (600 - 4Pir^3)/3Pir^2
Input this equal into the h in the cost equation, giving us;
C = 8Pir^2+2Pir((600 - 4Pir^3)/3Pir^2)
find the derivative of it and set it equal to zero. Your final derivative should be;
C' = 93Pir^3 - 3600
r should be 2.3 m, so 230 cm. Put 2.3 into the h equation to find h to be 8.96 m or 896 cm, which can be rounded to 900 cm.
The 16 m length of the tank is for the maximum and minimum, which we can ignore.
I hope this helps
This being a math help forum, I do not understand why you would post the solution to a problem.