STEP I 1990 question 1 solution

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Let . Then

(1)

Consider the lengths of the altitudes dropped

from Y to CD, from Y to AB and from X to AD.

They are

, and respectively.

Thus the areas of the triangles DCY, DYX, DXA are , and respectively, and the area of the quadrilateral is

(2)

When is fixed, and we wish to maximise this area:

Setting the derivative to zero, we get

Now, both and clearly belong in the interval (the latter because it is the angle marked ), and thus

which is the maximum. This can be seen by working out the second derivative,

which is clearly negative at .

Note that in this case, the maximum area is given by

(3)

If we wish to find the maximum when both and vary, we can proceed by finding the value of for which the is at a maximum. Thus,

Setting the derivative to zero:

Now, both and are in , and thus we can write

from which follows

(4)

Again, to make that that this is a maximum, let us take the second derivative:

which again can be seen to be negative, thus ensuring that (4) indeed gives us the maximum .

Now, insertion into (3) gives us the maximum area as

, i.e.

Just for clarification, note that this happens when

.

Alternate Solution

Now you could stop reading here, but for the extra educational benefit or something, I couldn't resist pointing out that this question follows very easily from Jensen's inequality, introduced to us avid STEP-solvers in 2007 II/7:

Let f(x) be a concave function on some interval and let lie in that interval. Jensen's inequality states that

and that equality holds if and only if

Now, in 2007 II/7, it was proven that was a concave function on .

Setting, in Jensen's inequality, , , , , and using (2) and (1) we get

with equality iff .

This immediately gives us the maximum.

Solution by ukgea.