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`        A Conical vessel is to be prepared out of a circular sheet of gold of unit radius. How much sectorial area is to be removed from the sheet so that vessel has maximum volume.`
6 months ago

```							hello praveen, we know that volume of cone V= pi*r^2*h/3also, r^2 + h^2= l^2, where l is the slant length, r is the radius of cone (not the circular sheet), h is cone’s height. also, l= radius of circular sheet = 1let the sector removed be of an angle x.so, the base circumference of the cone= 2pi*r, which must be equal to the circular sheet’s sector’s lengthso, 2pi*r= l*(x)= x (as l=1).also, h= sqrt(1 – r^2), so dh/dr= – r/halso, sectorial area removed= 1^2*x/2= x/2now, V= pi*r^2*h/3for max, dV/dx= 0so, (pi/3)[d(r^2*h)/dr]*(dr/dx)= 0but dr/dx = 1/2piso, d(r^2*h)/dr= 0or 2r*h + r^2*( – r/h)= 0or 2rh= r^3/hor h root2 = ror 2h^2= r^22 – 2r^2= r^2or r = root(2/3)but x= 2pi*rso, sectorial area removed= 1^2*x/2= x/2= pi*root(2/3) sq unitskindly approve :)
```
6 months ago
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