First Way
The problem asks for the minimum of the sum of two squares as shown below. If the sides of the squares are extended in our sketch to form a square of length AB on each side, four regions are formed:
Therefore, the minimum sum of the squares
occurs when the two rectangles have maximum area. But a rectangle has maximum area when it is a square or when AP = PB.
Second Way
Variations on the above approach include the following. Let AB = x and PB = y.
Then we want to minimize.
By the Arithmetic Mean-Geometric Mean inequality,
Therefore the sum of the two squares is always greater than the combined areas of the two rectangles except when x = y. So the minimum area occurs when P is the midpoint.
Third Way
Another approach is to formulate the area as a function of a single variable. Let AP = x and PB = AB - x. The the area is
This is a parabola with the following graph
where the vertex is at
.
Fourth Way
Another approach is the following, using the arithmetic mean -- geometric mean inequality.
with equality iff x = AB - xAnother . . .
For the high school student with just enough cookbook calculus to take f(x), find its derivative f'(x) = 4x - 2(AB), and set f'(x) = 0, the result is that he can conclude the function reaches the above minimum without having to think about it.
Still Another . . .
Another approach is to particularize the length AB and compute a sequence of values for sum of the two squares as P is placed along points on the line. Let AB = 10 and x = AP. Then the following table can be generated quickly.
x 0 1 2 3 4 5 6 7 8 9 10 10 - x 10 9 8 7 6 5 4 3 2 1 0 sum 100 82 68 58 52 50 52 58 68 82 100
This provides good intuition that the desired location for P is at the midpoint of AB.
And another . . .
Another variation is to construct a GSP animation to see the areas change as P is moved along AB.
