Evaluate $\min_{a,b \in [-1,1]}\max_{x\in [-1,1]}|(x-a)(x-b)|$.
Find maximas of $|(x-a)(x-b)|$ over $[-1,1]$. So I calculated the derivative and I finally found that this optimisation problem was the following : $$\min_{a,b \in [-1;1]} \max \left \{ \frac{1}{4}(a-b)^{2},(1+a)(1+b),(1-a)(1-b) \right \}$$
And now they ask me to resolve this using this graphic :
And here I have no idea how to proceed.Is the minimum at the 4 corners or is it at the intersection of the 3 planes ? Thanks
According to the graph the minimum is along the line $a+b=0$, one of the diagonals of the square $[-1,1]^2$. Hence, for $a,b\in [-1,1]$, $$\max \left \{ \frac{1}{4}(a-b)^{2},(1+a)(1+b),(1-a)(1-b) \right \}= \max \left \{ a^2,1-a^2 \right \}.$$ Therefore the minimum over $[-1,1]$ is attained when $a^2=1-a^2$, i.e. at $a=\pm1/\sqrt{2}$, the value there is $1/2$. $$\min_{a,b \in [-1,1]} \max \left \{ \frac{1}{4}(a-b)^{2},(1+a)(1+b),(1-a)(1-b) \right \}=\frac{1}{2}.$$ P.S. Note that the two minimum points $(\pm 1/\sqrt{2},\mp 1/\sqrt{2})$ are placed at the intersections of the three colored graphs (which are not planes).