170 3 Real Analysis
If a differentiable multivariable function has a global extremum, then this extremum
is found either among the critical points or on the boundary of the domain. We recall that
a point is critical if the (hyper)plane tangent to the graph is horizontal, which is equivalent
to the fact that all partial derivatives are equal to zero. Because any continuous function on
a compact domain attains its extrema, the global maximum and minimum exist whenever
the domain is closed and bounded. Let us apply these considerations to the following
problems.
Example.Find the triangles inscribed in the unit circle that have maximal perimeter.
Solution.Without loss of generality, we may assume that the vertices of the triangle have
the coordinates( 1 , 0 ),(coss,sins),(cost,sint),0≤s≤t≤ 2 π. We are supposed to
maximize the function
f(s,t)=√
(coss− 1 )^2 +(sins)^2 +√
( 1 −cost)^2 +(sint)^2
+√
(cost−coss)^2 +(sint−sins)^2
=√
2
(√
1 −coss+√
1 −cost+√
1 −cos(t−s))
= 2
(
sin
s
2+sin
t
2+sin
t−s
2)
,
over the domain 0≤s≤t≤ 2 π. To this end, we first find the critical points offin the
interior of the domain. The equation
∂f
∂s(s, t)=cos
s
2−cos
t−s
2= 0
gives coss 2 =cost− 2 s, and since boths 2 andt− 2 sare between 0 andπ, it follows that
s
2 =
t−s
2. The equation
∂f
∂t(s, t)=cost
2+cost−s
2= 0
implies additionally that coss=−coss 2 , and hences=^23 π. Consequently,t=^43 π,
showing that the unique critical point is the equilateral triangle, with the corresponding
value of the perimeter 3
√
3.
On the boundary of the domain offtwo of the three points coincide, and in that case
the maximum is achieved when two sides of the triangle become diameters. The value
of this maximum is 4, which is smaller than 3
√
- We conclude that equilateral triangles
maximize the perimeter.
502.Find the global minimum of the functionf:R^2 →R,