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How to show that it is impossible to have a cuboid for which the volume, the surface area and the perimeter are numerically equal ? The perimeter of a cuboid is the sum of the lengths of all its twelve edges.

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    $\begingroup$ Does your definition of "cuboid" require the faces to be rectangular? $\endgroup$ May 21, 2014 at 5:00

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I am assuming by "cuboid" you mean a rectangular box (which Wikipedia calls a "rectangular cuboid", the term "cuboid" being more general).

Then you are asking for a solution to the equations $$ 4(l + w + h) = 2(lw + wh + hl) = lwh. $$ Setting $c = l + w + h \in \mathbb{R}$, we construct the polynomial $$ (x - l)(x - w)(x - h) = x^3 - c x^2 + 2c x - 4c, $$ and we wish to know if it has three positive real roots for any real $c$. The discriminant of this polynomial is $$ -432 c^2+112 c^3-12 c^4 = -4c^2(3c^2 - 28c + 108), $$ which is strictly negative for all $c$, implying that there are two nonreal roots, and no solution to the equations exists for real numbers $l, w, h$.

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I found this question on TBO, and the official solution on it is:

We can quickly check that a cube cannot satisfy the conditions set, for surface area and perimeter to be equal the dimensions of the cube must be 2 and this does not give the correct result for volume. If we equate surface area and perimeter and rearrange we can show that at least one of the dimensions must be larger than two and less than 2. If we equate volume and surface area and rearrange we can show that each dimension must be larger than 2 which yields a contradiction. Hence the volume, surface area and perimeter of a cuboid cannot all be equal.

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    $\begingroup$ Anyone care to explain how rearrangement of the equation of the volume and the surface area implies that all dimensions are larger than 2? $\endgroup$
    – IChoi
    Dec 9, 2022 at 9:50

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