Q. 245.0( 2 Votes )

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)

2r6 = 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

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