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

OR

Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.

Given: A given quantity of metal is to be cast into a solid half circular cylinder with a rectangular base and semi-circular ends in order that total surface area is minimum.

To Prove: The ratio of length of cylinder to the diameter of semi-circular ends is ∶ + 2.

Let r is the radius and h is the height of the cylinder.

Volume of the cylinder, V =

⇒ h = ; k is constant parameter.

Total surface area, S = [2rh = bottom rectangle.]

⇒ S = rh + r² + 2rh

⇒ S =

⇒ S = ---- (a)

Differentiating with respect to r,

⇒

S to be minimum,

⇒

⇒

Putting the value of k in (a),

⇒

To find the value at which S is minimum, it is again differentiated with respect to r,

That means, at , S has a minimum value.

We also know,

So,

Putting the value of V,

[Proved]

OR

To Prove: Triangle of maximum area that can be inscribed in a given circle is an equilateral triangle

Let OA = OB = R be the radius of circle, AD = h be the height of triangle and BC = 2r be the base of the triangle.

OD is the perpendicular to the chord BC. That means, BD = DC = = r

Now, triangle OBD is a right-angle triangle. So,

OB² = BD² + OD²

R² = r² + (h – R)²

R² = r² + h² + R² – 2hR

r² = 2hR – h² --- (a)

We know, area of triangle,

A =

A² = r²h² = (2hR – h²) h² = 2h³R – h⁴

Differentiating with respect to h,

Area to be maximum,

Now,

Here putting ,

So, area is maximum for .

Putting in (i),

So, base of triangle, BC = 2r = R

Now, in right angle triangle ABD,

AB² = BD² + AD² = r² + h²

AB² =

AB = R

Similarly,

AC = R

So, AB = BC = AC = R

Hence, the triangle is equilateral. [Proved]