Henry Ernest Dudeney/Modern Puzzles/96 - Concerning a Cube/Solution

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Modern Puzzles by Henry Ernest Dudeney: $96$

Concerning a Cube
What is the length in feet of the side of a cube when
$(1)$ the superficial area equals the cubical contents;
$(2)$ the superficial area equals the square of the cubical contents;
$(3)$ the square of the superficial area equals the cubical contents?


Solution

$(1): \quad$ $6$ feet
$(2): \quad$ Approximately $1 \cdotp 57$ feet
$(3): \quad$ $\dfrac 1 {36}$ feet


Proof

Let $l$ denote the length of the side of a cube $\CC$.

Let $\AA$ denote the superficial area of $\CC$.

Let $\VV$ denote the cubical contents of $\CC$.

From Surface Area of Cube:

$\AA = 6 l^2$

From Volume of Cube:

$\VV = l^3$

Hence we can deduce the following:


Problem $(1)$

\(\ds \AA\) \(=\) \(\ds \VV\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds 6 l^2\) \(=\) \(\ds l^3\)
\(\ds \leadsto \ \ \) \(\ds l\) \(=\) \(\ds 6\)

$\Box$


Problem $(2)$

\(\ds \AA\) \(=\) \(\ds \VV^2\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds 6 l^2\) \(=\) \(\ds \paren {l^3}^2\)
\(\ds \) \(=\) \(\ds l^6\)
\(\ds \leadsto \ \ \) \(\ds l\) \(=\) \(\ds \sqrt [4] 6\)
\(\ds \) \(\approx\) \(\ds 1 \cdotp 565\)

$\Box$


Problem $(3)$

\(\ds \AA^2\) \(=\) \(\ds \VV\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds \paren {6 l^2}^2\) \(=\) \(\ds l^3\)
\(\ds \leadsto \ \ \) \(\ds l\) \(=\) \(\ds \dfrac 1 {36}\)

$\blacksquare$


Sources

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