Henry Ernest Dudeney/Puzzles and Curious Problems/225 - An Artist's Puzzle/Solution
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Puzzles and Curious Problems by Henry Ernest Dudeney: $225$
- An Artist's Puzzle
- An artist wished to obtain a canvas for a painting which would allow for
- the picture itself occupying $72$ square inches,
- a margin of $4$ inches on top and on bottom,
- and $2$ inches on each side.
- What is the smallest dimensions possible for such a canvas?
Solution
The picture will then be $6$ inches by $12$ inches.
Proof
Let $a$ inches be the width of the picture.
Let $A$ square inches be the total area of the canvas, including the picture and the borders.
We need to find $a$ such that $A$ is a minimum.
As the width is $a$, the height is $\dfrac {72} a$.
Hence:
| \(\ds A\) | \(=\) | \(\ds \paren {a + 4} \paren {\dfrac {72} a + 8}\) | a margin of $4$ inches on top and on bottom, and $2$ inches on each side. | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {a + 4} \paren {\dfrac {72} a + 8}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 72 + \dfrac {288} a + 8 a + 32\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 8 a + \dfrac {288} a + 104\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\d A} {\d a}\) | \(=\) | \(\ds 8 - \dfrac {288} {a^2}\) |
To make $A$ a minimum, we must make its derivative with respect to $a$ equal to zero.
Hence:
| \(\ds 8 - \dfrac {288} {a^2}\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2\) | \(=\) | \(\ds 36\) | simplification |
which leads us to the answer that the canvas must be $6 + 4 = 10$ inches wide and $\dfrac {72} 6 + 8 = 20$ inches high.
$\blacksquare$
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $225$. -- An Artist's Puzzle
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $275$. An Artist's Puzzle