An oil can is to have a volume 1000in^3 and is to be shaped like a sylinder with a flat bottom but capped by a hemisphere. Neglect the thickness of the material of the can and find the dimensions that will minimize the total amount of material needed to construct it.
Using Lagrange multipliers, you will need the following variables
radius, r
height, h
Volume, V=πr^2h+(2/3)πr^3=1000
Surface Area, A = πr^2+2πrh+2πr^2
λ=Lagrange multiplier
You need to minimize A subject to V=1000 using the following objective function:
Z(r,h,V,λ)
= πr^2+2πrh+2πr^2 + λ(V-1000)
find partial derivatives with respect to r, h, & λ to get 4 equations and solve for the system of equations in three unknowns r,h and λ.
Partial derivatives:
Zr=3r+h+L(r^2+rh)=0
Zh=2 %pi r + L %pi r^2 = 0
ZL=πr^2h+(2/3)πr^3-1000 =0
Eliminating L from first two equations gives r=h
Substitute h=r in third equation gives
r=(600/%pi)^(1/3)
=5.7588 approx.
Check my arithmetic
To minimize the total amount of material needed to construct the oil can, we need to consider the surface area of the can. We can break down the can into two components: the cylindrical part and the hemisphere cap.
Let's assume the radius of the cylinder is 'r' and the height of the cylinder is 'h'. The volume of the cylinder is given as 1000 in³.
We know that the volume of a cylinder is given by the formula V = πr²h.
So, V = 1000 in³. We can rearrange this equation to solve for h:
h = 1000 / (πr²).
Let's consider the surface area of the cylinder first. The lateral surface area of a cylinder is given by the formula A = 2πrh. Since the bottom of the can is a flat surface, it doesn't contribute to the surface area.
Now, let's move on to the hemisphere cap. The surface area of a hemisphere is given by the formula A = 2πr². However, since the hemisphere is being used as a cap, we only need half of the surface area, so the contribution is πr².
To minimize the total amount of material, we need to minimize the sum of the surface areas of the cylinder and the hemisphere cap.
Let's denote the total surface area of the oil can as S:
S = 2πrh + πr².
Since we want to minimize S, we need to minimize it with respect to the variables r and h. To do this, we can perform partial derivatives.
∂S/∂r = 2πh + 2πr,
∂S/∂h = 2πr.
To find the critical points, we set the partial derivatives equal to zero and solve the resulting system of equations:
2πh + 2πr = 0, and
2πr = 0.
Solving this system, we find that h = 0, which means the bottom of the oil can would have zero height. Since this is not physically possible, it is not a valid solution.
Therefore, there are no critical points for S within the given constraints.
However, in practical terms, we can still find dimensions that will minimize the total amount of material needed. We can make an observation: the smaller the radius and the greater the height, the smaller the amount of material used. So, the dimensions that will minimize the total amount of material would be the smallest possible radius and the largest possible height that still maintain a volume of 1000 in³.
Therefore, the dimensions that will minimize the total amount of material needed to construct the oil can would be a cylinder with a very small radius and a very large height.