Maximize the Area
This problem is from the newsletter for users of the TI graphing calculators.
Given a rectangular piece of paper, label the four corners of the paper A, B, C, and D (see figure below). Fold the vertex A so that it just touches the segment CD. Determine the location of point E that maximizes the area of the triangle EDF.
Please submit all work for this problem to mathpotw@acu.edu no later than 5:00 PM on Thursday, September 26th.
Solution to Maximize the Area
Correct solutions were submitted by: Bethany Witemeyer.
We let the distance from A to D be 1 and the distance from F to D be denoted by x. It follows that the distance from F to E is 1-x, since it is equal to the distance from A to F. Therefore we have a right triangle with height x and hypotenuse of length 1-x. Thus the square of the length of the base is (1-x)x – x2= 1-2x. We want to find x that maximizes the area of this triangle.
It will be easier if we maximize the square of the area since we then have no square roots to differentiate. Thus we differentiate AREA2 = x2(1-2x)/4 with respect to x and set the result equal to zero. Since x = 0 would give us a “triangle” with area zero, we know that x is not zero. So the solution is x = 1/3. Therefore the distance from D to E is SQRT(3)/3.
Another solution: A momentary contemplation allows one to observe that the triangle of maximum area is a 30-60-90-degree triangle. It makes one conjecture that there is a geometric proof not using calculus. Indeed, Alexander Karabegov submitted such a solution. He merely reflected the triangleDFE through the segment DE to obtain an isosceles triangle with sides of length 2x, 1-x, and 1-x. The sum of the three sides is 2 (independent of x), so we see that we are attempting to find an isoceles triangle of maximum area with fixed perimeter 2. Some persons know that such a triangle is an equilateral triangle, hence the same result as before.
