A interesting max min problem Let $\mathcal{S}\subset\mathbb{R}^2$ be a bounded, closed, compact, convex set which contains origin in its interior.
Define 
\begin{align}
c_1^{\star}=\min_{{(x_1,0)\in\mathcal{S}}}~&x_1
\end{align}
Also define
\begin{align}
\lambda(t)&=\min_{(x_1,x_2)\in\mathcal{S}}~~x_1+tx_2 \\
c_2^{\star}&=\max_{t}\lambda(t)
\end{align}
Then is it true that 
\begin{align}
\lambda(t) \leq c_1^{*}
\end{align}
irrespective of choice of $t$.
Also, when is the following true
\begin{align}
c_2^{\star}= c_1^{\star}
\end{align}
 A: Yes, it is true that $\forall t \lambda (t) \le c_1^*$ because one candidate point for $\lambda(t)$ is $(x_1,0)$ at which point $x_1+tx_2=x_1$, so minimizing over $S$ must be less than or equal to this.  One candidate value for $c_2^*$ is $\lambda(0)=c_1^*$.  Since $c_2^*$ is the maximum over $t$, it can't be less than this.  If $S$ is an axis aligned rectangle that straddles the $x$ axis, $c_1^*$ will be the $x$ coordinate of the left edge.  For $t \gt 0$, $\lambda(t)$ will be obtained at the lower left corner and $\lambda(t) \lt \lambda(0)$.  For $t \lt 0$, $\lambda(t)$ will be obtained at the upper left corner and again less than $\lambda(0)$.  So $c_2^*=\lambda(0)=c_1^*$  
Added:  Yes, it is true for all compact $S$ that include at least one point of the $x$ axis (not necessarily convex, nor necessarily including the origin in the interior) that $c_2^*=c_1^*$.  Because $S$ is compact, the intersection of $S$ and the $x$ axis is closed and there is a leftmost point in the intersection.  This point is $(c_1^*,0)$.  Let us define $L(x_1,x_2,t)=x_1+x_2t$ for $(x_1,x_2) \in S$.  Then $\lambda(t)=\min_{(x_1,x_2)\in S}L(x_1,x_2,t)$  Clearly $\forall t\lambda(t) \le L(c_1^*,0,t)=c_1^*$, so $c_2^* \le c_1^*$.  We showed above that $c_2^* \ge c_1^*$, so $c_2^*=c_1^*$
