The
excellent book Challenging Problems in Algebra, by Alfred Posamentier
and Charles T. Salkind, has this problem: Return to my Puzzle pages
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© Copyright 2004, Jim Loy
The
excellent book Challenging Problems in Algebra, by Alfred Posamentier
and Charles T. Salkind, has this problem:
From a rectangular cardboard 12x14, an isosceles trapezoid and a square of side length s are removed so that their combined area is a maximum. Find the value of s.
Now this seems a little ambiguous to me. Maybe the trapezoid has a short base of s, maybe not. The solution in the book is accompanied by the diagram above left, which finally clears things up somewhat. Using that diagram, and any of a number of methods, we find that s = 12 and the area of the trapezoid = 0.
I don't see why we can't use the second diagram above, where the solution is s = 12, and the area of the trapezoid (which has become a rectangle) = 24.
Problems like these are solved by finding an expression for the total area (of square + trapezoid above), or some other quantity. As we change the lengths (s above), we find that the total area either increases or decreases or increases for a while and then decreases. If it simply increases or decreases, then our solution is at one endpoint of the range. However, if the area increases and then decreases, we know from calculus that the max (or min) occurs when the Derivative (slope) of our expression is zero (the graph is horizontal), assuming that the graph is smooth. Normally, the derivative is easy to find.
An
example of such a problem is this: We want to build a fence, one mile long,
that "encloses" a maximal rectangular area, so that the fence makes up three
sides of the rectangle and a straight river bank makes up the fourth side (see
the diagram on the right).
We
find that when a = 0, the area = 0, and when a = 1/2, the area = 0. Maybe the
solution is a square, a = 1/3 and area = 1/9. Using symmetry, as the expression
for the area graphs as a parabola (see the graph on the left), and a parabola
is symmetric about its maximum (when oriented as in this case), we find that
the solution is a = 1/4 and the area = 1/8. Instead of symmetry, we could have
used the derivative.
If there were no river, and the fence made up all four sides of our rectangle, then we would find that a square is the maximal rectangle for a given fixed perimeter, which confirms what we probably already knew.