566 Real Analysis
We want to maximize the expressionab
√
1 −x^2 +cd√
1 −y^2 , which is twice the area
of the rectangle. Let
f (x, y)=ab√
1 −x^2 +cd√
1 −y^2 ,
g(x, y)=a^2 +b^2 − 2 abx−c^2 −d^2 + 2 cdy.We are supposed to maximizef (x, y)over the square[− 1 , 1 ]×[− 1 , 1 ], with the con-
straintg(x, y)= 0. Using Lagrange multipliers we see that any candidate for the
maximum that lies in the interior of the domain satisfies the system of equations
−ab2 x
√
1 −x^2=−λ 2 ab,−cd
2 y
√
1 −y^2=λ 2 cd,for someλ. It follows that
√
1 −x^2 /x=−√
1 −y^2 /y, and so the tangents of the opposite
angles are each the negative of the other. It follows that the angles are supplementary.
In this casex =−y. The constraint becomes a linear equation inx. Solving it and
substituting in the formula of the area yields the Brahmagupta formula
A=√
(s−a)(s−b)(s−c)(s−d), wheres=a+b+c+d
2.
Is this the maximum? Let us analyze the behavior offon the boundary. Whenx= 1
ory=1, the quadrilateral degenerates to a segment; the area is therefore 0. Let us see
what happens wheny=−1. Then the quadrilateral degenerates to a triangle, and the
area can be computed using Hero’s formula
A=√
s(s−a)(s−b)(s−(c+d)).Sinces(s−(c+d))<(s−c)(s−d), we conclude that the cyclic quadrilateral maximizes
the area.
(E. Goursat,A Course in Mathematical Analysis, Dover, New York, 1904)
511.Without loss of generality, we may assume that the circle has radius 1. Ifa, b, c
are the sides, andS(a, b, c)the area, then (because of the formulaS=pr, wherep
is the semiperimeter) the constraint readsS=a+b 2 +c. We will maximize the function
f (a, b, c)=S(a, b, c)^2 with the constraintg(a, b, c)=S(a, b, c)^2 −(a+ 2 b+c)^2 =0.
Using Hero’s formula, we can write
f (a, b, c)=
a+b+c
2·
−a+b+c
2·
a−b+c
2·
a+b−c
2
=−a^4 −b^4 −c^4 + 2 (a^2 b^2 +b^2 c^2 +a^2 c^2 )
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