The base of prism is equilateral triangle. The distance from the centre of one base to one of the vertices of the other base is `ldot` Then altitude of the prism for which the volume is greatest is `l/2` (b) `l/(sqrt(3))` (c) `l/3` (d) `l/4`

Options

A. `(l)/(2)`
B. `(l)/(sqrt(3))`
C. `(l)/(3)`
D. `(l)/(4)`

Answer

2

Explanation

From the figure `l^(2)=h^(2)+x^(2)`
Area of base (equilateral triangle) is `(sqrt(3))/(4)a^(2)`
Also `(3x)/(2) =a sin 60^(@)=(sqrt(3))/(2)a`
`therefore a=sqrt(3)x`
Now volume (v) = height `xx` area of base ltbr. `=h(sqrt(3))/(4)a^(2)=h(sqrt(3))/(4).3x^(2)`
`=(3sqrt(3))/(4)h(l^(2)-h^(2))`
`=(3sqrt(3))/(4) (hl^(2)-h^(2))`
`therefore (dV)/(dh)=(3sqrt(3))/(4)(l^(2)-3h^(2))`
`(dv)/(dh)=0rarr(l)/(sqrt(3))`which is point of maxima