146 CHAPTER 3 Inequalities
-4 -2 2 4-4-224
x +y =4 2xy=c>0xx2 2
x
y
2xy=c< 0
Problem. Given that x^2 +y^2 = 4,
find the maximum value of 2xy.
Solution. If we are thinking in
terms of the above-mentioned in-
equality 2xy≤x^2 +y^2 , with equality
if and only ifx=y, then we see im-
mediately that the maximum value of
2 xymust bex^2 +y^2 = 4. However, it
is instructive to understand this prob-
lem in the context of the graph to the
right, where the “constraint curve” is
the graph ofx^2 +y^2 = 4 and we’re trying to find the largest value of
the constantcfor which the graph 2xy=cmeets the constraint curve.
From the above figure, it is clear that where 2xy obtains its maxi-
mum value will occur at a point where the graph is tangent to the circle
with equationx^2 +y^2 = 4. As a result, this suggest that the solution
can also be obtained using the methodology of differential calculus (in-
deed, it can!), but in this chapter we wish to stress purely algebraic
techniques.
We can vary the problem slightly and ask to find the maximum value
ofxy given the same constraintx^2 +y^2 = 4. However, the maximum
ofxyis clearly 1/2 the maximum of 2xyand so the maximum value of
xyis 2.
In an entirely similar fashion we see that the minimum value of 2xy
givenx^2 +y^2 = 4 must be−2. This can be seen from the above figure.
Even more elementary would be to apply the inequality 0≤(x+y)^2 ⇒
− 2 xy≤x^2 +y^2.