先把位于凸包的点求出，然后n^2枚举每两个点x,y，接着左右边找个离线最远的点。
可以知道，当x不变y单调递增时，两边距离最远的两点也在单调递增。
于是可以使用旋转卡壳。
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <cctype>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define down(i, l, r) for(int i=l; i>=r; i--)
#define maxn 2009
#define pi acos(-1)
#define eps 1e-11
using namespace std;
inline int read()
{
int x=0, f=1; char ch=getchar();
while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();
return x*f;
}
struct P{double x, y;} p[maxn], s[maxn];
P operator - (P a, P b){return (P){a.x-b.x, a.y-b.y};}
double operator * (P a, P b){return a.x*b.y-a.y*b.x;}
double dis(P a, P b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}
bool cmp(P a, P b)
{
double t=(a-p[1])*(b-p[1]);
if (t==0) return dis(a, p[1])<dis(b, p[1]); else return t<0;
}
int n, top;
double ans;
int main()
{
n=read();
rep(i, 1, n) scanf("%lf%lf", &p[i].x, &p[i].y);
int k=1;
rep(i, 2, n) if (p[i].y<p[k].y || (p[i].y==p[k].y && p[i].x<p[k].x)) k=i;
swap(p[1], p[k]);
sort(p+2, p+n+1, cmp);
s[++top]=p[1]; s[++top]=p[2];
rep(i, 3, n)
{
while (top>1 && (p[i]-s[top-1])*(s[top]-s[top-1])<=0) top--;
s[++top]=p[i];
}
s[top+1]=p[1];
int a, b;
rep(x, 1, top)
{
a=x%top+1; b=(x+2)%top+1;
rep(y, x+2, top)
{
while (a%top+1!=y && (s[y]-s[x])*(s[a+1]-s[x])>(s[y]-s[x])*(s[a]-s[x])) a=a%top+1;
while (b%top+1!=x && (s[b+1]-s[x])*(s[y]-s[x])>(s[b]-s[x])*(s[y]-s[x])) b=b%top+1;
ans=max((s[y]-s[x])*(s[a]-s[x])+(s[b]-s[x])*(s[y]-s[x]), ans);
}
}
printf("%.3lf
", ans/2);
return 0;
}