please answer in detailSolution112 max fxy 1 y2 and min of

please answer in detail

1.12) max f(x,y) = 1- y^2. and min of (max f(x,y)) = 0.

so v+ = 0.

min f(x,y) for y = x^2 - 1. max( min(f(x,y)) = 0 taking x=1.

so v- =0.

1.13) max using x for f(x,y) = (1-y)^2. and min(max f(x,y)) = 0.

so v+ =0

similarly min using y for f(x,y) = (x-1)^2 and max(min(f(x,y)) = 4 at x = -1.

1.16) for all y belongs to D.

min f(x*,y) using y = v

Now for all x the value <= v.

So atleast one value pair(x,y) is present such that f(x*,y*)=v

Hence max min f(x,y) = v.

so f(x*,y*) =max(min(f(x,y)) = min(max(f(x,y)) = v.