Given a cylinder of fixed volume V, the total surface area (including the two circular ends) is minimized for a radius of R and height H. Find H/R.
Given a cylinder of fixed volume V, we need to find the height-to-radius ratio (H/R) that minimizes the total surface area.

Let's start by expressing the volume and surface area in terms of the cylinder's radius and height:

- The volume of the cylinder is given by V = πR^2H, where R is the radius and H is the height.
- The total surface area (including the top and bottom circular ends) is given by A = 2πR^2 + 2πRH.

To minimize A subject to the constraint on V, we can use the method of Lagrange multipliers or calculus. However, in this case, we can apply the AM-GM inequality to obtain a neat solution without differentiation.

By AM-GM, we have:

A = 2πR^2 + 2πRH
  = 2πR^2 + πRH + πRH
  ≥ 3(2πR^2)(πRH)(πRH)^(1/3)  (since (2πR^2)^(1/3)≥0 and (πRH)^(2/3)≥0)
  = 3(2π^3R^4H^2)^(1/3).
  
Note that the inequality becomes equality when the two terms (2πR^2) and (πRH) are equal, which implies H/R = 2.

Since V = πR^2H, R^2H=V/π. Substituting this into the expression for A, we get:

A = 3(2πV^2)^(1/3) / H^(1/3)

To minimize A subject to the constraint V = constant, we need to minimize the right-hand side, which is decreasing in H. Therefore, the minimum occurs when H is as large as possible, which is achieved when H/R is 2.

Hence, the solution is H/R = 2, meaning that the height is twice the radius for the cylinder of fixed volume that minimizes the total surface area.
Complete the following task based on the instruction and optional input provided.### Instruction:
Given a cylinder of fixed volume V, the total surface area (including the two circular ends) is minimized for a radius of R and height H. Find H/R.

### Output:
Given a cylinder of fixed volume V, we need to find the height-to-radius ratio (H/R) that minimizes the total surface area.

Let's start by expressing the volume and surface area in terms of the cylinder's radius and height:

- The volume of the cylinder is given by V = πR^2H, where R is the radius and H is the height.
- The total surface area (including the top and bottom circular ends) is given by A = 2πR^2 + 2πRH.

To minimize A subject to the constraint on V, we can use the method of Lagrange multipliers or calculus. However, in this case, we can apply the AM-GM inequality to obtain a neat solution without differentiation.

By AM-GM, we have:

A = 2πR^2 + 2πRH
  = 2πR^2 + πRH + πRH
  ≥ 3(2πR^2)(πRH)(πRH)^(1/3)  (since (2πR^2)^(1/3)≥0 and (πRH)^(2/3)≥0)
  = 3(2π^3R^4H^2)^(1/3).
  
Note that the inequality becomes equality when the two terms (2πR^2) and (πRH) are equal, which implies H/R = 2.

Since V = πR^2H, R^2H=V/π. Substituting this into the expression for A, we get:

A = 3(2πV^2)^(1/3) / H^(1/3)

To minimize A subject to the constraint V = constant, we need to minimize the right-hand side, which is decreasing in H. Therefore, the minimum occurs when H is as large as possible, which is achieved when H/R is 2.

Hence, the solution is H/R = 2, meaning that the height is twice the radius for the cylinder of fixed volume that minimizes the total surface area.