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 -
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 –
⇒ 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.
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.
As we need to maximise A.
For A to be minimum,
∴ Differentiating equation 2 w.r.t x we get –
⇒ 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 –
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