In a regular triangular prism,the distance fr | vedclass

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(B) Let $a$ be the side length of the equilateral triangular base and $h$ be the altitude of the prism.The distance from the centre $G$ of the base to

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$\frac{l}{\sqrt{3}}$

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(B) Let $a$ be the side length of the equilateral triangular base and $h$ be the altitude of the prism.The distance from the centre $G$ of the base to a vertex $A$ is the circumradius of the equilateral triangle,given by $R = \frac{a}{\sqrt{3}}$.In the right-angled triangle formed by the altitude $h$,the circumradius $R$,and the distance $l$,we have $l^2 = R^2 + h^2 = \frac{a^2}{3} + h^2$.Thus,$a^2 = 3(l^2 - h^2)$.The volume of the prism is $V = 	ext{Area of base} 	imes h = \left( \frac{\sqrt{3}}{4} a^2 ight) h$.Substituting $a^2$,we get $V(h) = \frac{\sqrt{3}}{4} 	imes 3(l^2 - h^2) h = \frac{3\sqrt{3}}{4} (l^2 h - h^3)$.To maximize $V$,we find $V'(h) = \frac{3\sqrt{3}}{4} (l^2 - 3h^2) = 0$.Setting $V'(h) = 0$ gives $l^2 = 3h^2$,so $h = \frac{l}{\sqrt{3}}$.Thus,the altitude for which the volume is greatest is $\frac{l}{\sqrt{3}}$.

In a regular triangular prism,the distance from the centre of one base to one of the vertices of the other base is $l$. Find the altitude of the prism for which the volume is greatest.

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