Show that the rig

Let, r be the radius of the right-circular cone, l be the slant height, and h be the altitude of a given right circular cone.

Given, that volume of cone (V = constant) and curved surface area of cone(C) = minimum

To prove: h = √2r

∵ curved surface area of cone = πrl

∴ C = πrl

∵ l can determined by using Pythagoras theorem.

∴ C = = πr√(r²+h²) …(1)

∵ Volume is constant.

So we can relate the height and radius with the help of V.

As we know that -

⇒ …(2)

Using equation 1 and 2, we can write –

As we need to minimise C,

If C is minimum C² will also be minimum, and we can also say that converse is true.

So, it is easy to minimise C² term

∴ squaring both sides, we get –

⇒

For S to be minimum,

∴ Differentiating w.r.t r we get –

⇒ …(3)

∴

⇒ 4π²r⁶ = 18V²

⇒

⇒

⇒ 2r⁶ = r⁴h²

⇒ h² = 2r²

⇒ h = √2r

Now, we need to check the sign of

∴ differentiating equation 3 again w.r.t r –

⇒

∴

⇒

∴ S is minimum, or curved surface area of a right circular cone is minimum for a given volume at h = √2r.

OR

Let x and y be the length and breadth of a rectangle, and an equilateral triangle is surmounted over it.

∴ the side of equilateral triangle = x

Given the perimeter of the window = 12 m

⇒ x + 2y + 2x = 12

⇒ 3x + 2y = 12

⇒ 2y = 12 – 3x

⇒ y = 6 – (3/2)x …(1)

Let A denotes the area of the window.

∴

⇒

⇒ …(2)

As we need to maximise A.

For A to be minimum,

∴ Differentiating equation 2 w.r.t x we get –

⇒ …(3)

⇒

⇒ 12 - 6x + √3x = 0

⇒

Putting the value of x in equation 1 we get-

∴ Differentiating equation 3 again, we get –

∴ A is maximum or area of the window is maximum for a given dimension of the rectangle –

and