Henry Ernest Dudeney/Puzzles and Curious Problems/225 - An Artist's Puzzle/Solution

by : $225$

 * An Artist's Puzzle

Solution

 * $10$ inches wide by $20$ inches high.

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:

To make $A$ a minimum, we must make its derivative $a$ equal to zero.

Hence:

which leads us to the answer that the canvas must be $6 + 4 = 10$ inches wide and $\dfrac {72} 6 + 8 = 20$ inches high.