Q. 245.0( 2 Votes )

Show that the rig

Answer :

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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