The container consists of three parts. A half sphere on the top, a cylindrical in the middle and another half sphere on the bottom. The volume of the container must be 1.5 m³. The height of the cylindrical is 1 m. What is the radius of the container?

Let the height of the cylindrical be h, and let the volume and the radius be v and r respectively.

The volume, v, of the container is the sum volume of the cylindrical and the two hemisphere. It can be expressed by

     v = πr²h + 2(3πr³/8) = πr²h + 3πr³/4

Substituting h = 1 m into the above equation gives

     v = πr² + 3πr³/4

The function v = πr² + 3πr³/4 is plotted in the diagram on the left. This curve intersects the line v = 1.5 around 0.65. Start Newton's Method to solve the equation v(r) =1.5 with r₁ = 0.65 as the starting point. First, calculate the derivative of the volume, v, with respect to the radius, r

     f ^'(r) = dv/dr

          = d(πr² + 3πr³/4)/dr

          = d(πr²)/dr + d(3πr³/4)/dr

          = 2πr + 9πr²/4

Newton's Method gives x₂ = x₁ - f(x₁)/ f '(x₁). When r₁ = 0.65, v and f ^'(r) can be calculated.

     v₁ = πr₁² + 3πr₁³/4 -1.5

       = π(0.65)² + 3π(0.65)³/4 -1.5

       = 0.474

     f ^'(r₁) = 2πr₁ + 9πr₁²/4

           = 7.071

r₂ can be found with Newton's Method as

     r₂ = r₁ - v₁/ f '(r₁)

        = 0.65 - 0.14/6.315

        = 0.583

Now use r₂ = 0.583 as the starting point for the second iteration.

     v₂ = πr₂² + 3πr₂³/4 -1.5

       = π(0.583)² + 3π(0.585)³/4 -1.5

       = 0.035

     f ^'(r₂) = 2πr₂ + 9πr₂²/4

           = 6.066

r₃ can be found as

     r₃ = r₂ - v₂/ f '(r₂)

        = 0.583 - 0.035/6.066

        = 0.577 m

More iterations can be processed, however, the result will be closer and closer to 0.577 m. For the vessel to hold 1.5 m³, its radius must be about 0.577 m.