Prove that the maximum volume of a rectangular parallelepiped with a fixed surface area is when it is a cube.(adsbygoogle = window.adsbygoogle || []).push({});

Volume = v, length = l, width = w, height = h and surface area = s. Then we have: v = l*w*h = h^3. s = 2*w*l + 2*w*h + 2*h*l = constant = 6*h^2.

For a maximum to exist [tex] (\frac{{\partial ^2 y}}{{\partial x^2 }})(\frac{{\partial^2 y}}{{\partial x^2}}) - (\frac{{\partial^2 y}}{{\partial x}{\partial t }}) [/tex] > 0.

Where x, y and t are unknown to me. I just need some help to get started. Thank you very much.

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# Homework Help: Maximum volume question

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