# Prove that the maximum volume of a triangular-base prism is $\sqrt{\dfrac{K^3}{54}}$ where K is the area of three triangles containing a vertex A

Consider a prism with triangular base. The total area of the three faces containing a particular vertex $A$ is $K$. Show that the maximum possible volume of the prism is $\sqrt{\frac{K^3}{54}}$ and find the height of this largest prism.

I have no idea how to approach the problem. Please help. I know we need to use the properties of triangles and also the AM-GM inequality somewhere, but cannot put it together to solve the problem.

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Is it a prism or a pyramid? Your title seems to imply you mean pyramid - "three triangles containing a vertex A" but the post implies prism, "The total area of the three faces containing a particular vertex". If it's a prism, then only one face is a triangle, the other two are rectangles, correct? – Bennett Gardiner May 2 '14 at 4:40
@BennettGardiner A prism with triangular base, isn't that a pyramid? Maybe I am wrong. – Hawk May 2 '14 at 4:42
Prism... Pryamid – Bennett Gardiner May 2 '14 at 4:43
@BennettGardiner I hope the edited version is correct. – Hawk May 2 '14 at 4:47
You still haven't indicated what shape you meant. Did you click on the links and look at the google images? – Bennett Gardiner May 2 '14 at 5:15