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# In a regular triangular prism the distance from the centre of one base to one of the vertices of the otherbase is l . The altitude of the prism for which the volume is greatest

Jitender Singh IIT Delhi
7 years ago
Ans:
Let the altitude of th prism to be ‘h’ & height of triangle part of the prism to be ‘H’.
Given:
$(\frac{H}{2})^{2}+(\frac{h}{2})^{2} = l^{2}$
$H = 2\sqrt{l^{2}-(\frac{h}{2})^{2}}$
Area of triangle:
$A = \frac{H^{3}}{\sqrt{3}} = \frac{4}{\sqrt{3}}(l^{2}-\frac{h^{2}}{4})$
Volume of Prism ‘V’ :
$V = \frac{1}{2}Ah = \frac{4h}{\sqrt{3}}(l^{2}-\frac{h^{2}}{4})$
To maximise the volume,
$\frac{\partial V}{\partial h} = \frac{4}{\sqrt{3}}(l^{2}-\frac{3h^{2}}{4}) = 0$
$\Rightarrow h = \frac{2l}{\sqrt{3}}$
Thanks & Regards
Jitender Singh
IIT Delhi