Question

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semi - circular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its semi - circular ends is π : (π + 2).

Answer 1

Let ‘h’ be the height ‘or’ length of half cylinder, ‘r’ be the radius of half cylinder and ‘d’ be the diameter. 

We know that, 

⇒ Volume of half cylinder (V) = \(\frac{1}{2}\)πr²h

⇒ h = \(\frac{2V}{\pi r^2}\) ...(1)

Now we find the Total surface area (TSA) of the half cylinder, 

⇒ TSA = Lateral surface area of the half cylinder + Area of two semi - circular ends + Area of the rectangular base

We need total surface area to be minimum and let us take the TSA as the function of r, 

For maxima and minima,

Differentiating TSA again,

We have got,

Total surface area minimum for r = \(\frac{(\pi+2)h}{2\pi}\)

We know that, 

Diameter is twice of radius

∴ Thus proved.