1.

What is the value of 6t such that volume contained inside the planes sqrt(1-t^(2))x+tz=2sqrt(1-t^(2)), z=0,x=2+(tsqrt(4t^(2)-5t+2))/(sqrt(12(1-t^(2))^(1//4))) and |y|=2 is maximum.

Answer»


Solution :All the given planes are at right to `XZ` plane so a across section paallel to `zx` plane will be same every where. So, volume will be maximum when area of triangular cross section is maximum.
When we cut these planes by the plane `y=0`, the triangle OBTAINED is `ABC`, where
`AB=(tsqrt(4t^(2)-5+2))/(SQRT(12)(1-t^(2))^(1//4))`
LET `CAB=theta=` angle between first and `xy` plane, the `costheta=t`
Area of `DeltaABC=Delta=1/2(AB)^(2)tantheta`
`Delta=1/2(t^(2)(4t^(2)-5t+2))/(12(1-t^(2))^(1/2))XX(sqrt(1-t^(2)))/t=1/2 (4t^(3)-5t^(2)+2t)/12`
For example `(dDelta)/(dt)=0implies12t^(2)-10t+2=0`
`impliest=(10+-sqrt(100-96))/24=1/2` or `1/3`
Now, `(d^(2)Delta)/(dt^(2))=24t-10`
So, `Delta`"is maximum at"`t=1/3`"and minimum at "`t=1/2`
`:. t=1/3`


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