सिद्ध कीजिए की दी हुई तिर्यक ऊंचाई और महत्तम आयतन वाले शंकु का अर्ध शीर्ष कोण `tan^(-1)sqrt(2)` होता है |

Options

A. `2`
B. `1`
C. `sqrt(2)`
D. `sqrt(3)`

Answer

C

Explanation

Volume of cone `V=(pi)/3 r^(2)h`
`impliesV=(pi)/3 r^(2) sqrt(l^(2)-r^(2))`
On differentiating both sides w.r.t `r` we get
`(dV)/(dr)=(pi)/3 [2rsqrt(l^(2)-r^(2))+(r^(2))/(2sqrt(l^(2)-r^(2)))(-2r)]`
Put `(dV)/(dr)=0`

`implies2r(sqrt(l^(2)-r^(2)))-(r^(3))/(sqrt(l^(2)-r^(2)))=0impliesr[2(l^(2)-r^(2))-r^(2)]=0`
`implies 2l^(2)-3r^(2)impliesr=+-lsqrt(2/3)`
At `r=lsqrt(2/3),(d^(2)V)/(dr^(2))lt0,` maxima
`:. h=sqrt(l^(2)-2/3l^(2))=l/(sqrt(3))`
In `DeltaABC, tan theta=r/h=(lsqrt(2/3))/(l/(sqrt(3))=sqrt(2)`