I cannot find a way to define a function as the minimum of two other functions that each have the same two independent variables. I have f(x,y) and g(x,y), and I would like to find $h = \min\{f(x,y),g(x,y)\}$. $h$ is not a numerical answer, but a piecewise function; i.e., $h = f(x,y)$ for $f(x,y)<g(x,y)$, and $h=g(x,y)$ for $f(x,y)>=g(x,y)$.

Additionally, I would like to find the function that maximizes $h$ with respect to the variable $y$ and plot it in the xy-plane.

  • $\begingroup$ Sounds like homework. $\endgroup$
    – Alan
    May 1, 2018 at 0:18
  • $\begingroup$ What have you tried? Have you looked at Min in the documentation? $\endgroup$
    – eyorble
    May 1, 2018 at 0:22
  • 1
    $\begingroup$ I am doing an analysis on a baseball bunting play where x is the distance the fielder is from the plate and y is the angle the ball is hit at. G and F represents the likelihood of an out given that a specific player fields the ball. Thus it is natural that the minimum of these functions would indicate who should field the ball. While the hitter maximizes this likelihood of a hit fcn, h, by choosing the angle y. $\endgroup$ May 1, 2018 at 1:10
  • $\begingroup$ Under Min in documentation it is not used as function, rather an operation. $\endgroup$ May 1, 2018 at 1:11

1 Answer 1


Since you haven't said what f and g are, I've just picked something at random:

f[x_, y_] := x y;
g[x_, y_] := x^2 + y^2;
h[x_, y_] := Min[f[x, y], g[x, y]]

If you prefer, you can explicitly make h a piecewise function:

h2[x_, y_] := Piecewise[{{f[x, y], f[x, y] <= g[x, y]},                      
                         {g[x, y], f[x, y] > g[x, y]}}]
  • 1
    $\begingroup$ For fun: h3 = Through@*Min[f, g]. $\endgroup$ May 1, 2018 at 3:28
  • $\begingroup$ h[x_, y_] = Min[f[x, y], g[x, y]] // PiecewiseExpand $\endgroup$
    – Bob Hanlon
    May 1, 2018 at 5:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.