1962 AHSME Problems/Problem 37
is a square with side of unit length. Points and are taken respectively on sides and so that and the quadrilateral has maximum area. In square units this maximum area is:
Let Or As So Equality occurs when So the maximum value is
Let us first draw a unit square. We will now pick arbitrary points and on and respectively. We shall say that Thus, our problem has been simplified to maximizing the area of the blue quadrilateral. If we drop an altitude from to , and call the foot of the altitude , we can find the area of by noting that . We can now finish the problem.
Since and , we have:
To maximize this, we compete the square in the numerator to have:
Finally, we see that , as: So,
where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by , which does not change the inequality sign. Thus, the maximum area is or when
Solution 3 (Calculus)
The area of the quadrilateral can be expressed as .
By taking the derivative of , we get .
We make and get the critical point .
Substituting , the maximum area is or .
Plot the points of the quadrilateral (with the bottom left corner of the square being at the origin) to plug into shoelace formula: So, Trying first, you get . Any larger or smaller value gives an answer less than . Therefore, the answer is .