2 added 56 characters in body edited Aug 2 '14 at 5:17 unlikely 4,6401414 silver badges3939 bronze badges Too long for a comment but... For your specific setup this can be a way with a v10 function: tt1 = Table[Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5, 1}, {y, 0, 5, 1}] With[{v = z /. tt1}, With[{m = Max[v]}, {m, Most@FirstPosition[v, Max[v]]m]}]]]  But... whyWhy don't you solesolve analitically the problem? The command z /. Solve[z^2 == x^2 y - z, z]  gives {1/2 (-1 - Sqrt[1 + 4 x^2 y]), 1/2 (-1 + Sqrt[1 + 4 x^2 y])}  so it's not surprising what and where is the maximum is located at lower-right corner of the matrix, becausewhere $$x=y=5$$, and here %[] /. {x -> 5, y -> 5}  gives: 1/2 (-1 + Sqrt)  Too long for a comment but... For your specific setup this can be a way tt1 = Table[Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5, 1}, {y, 0, 5, 1}] With[{v = z /. tt1}, {Max[v], Most@FirstPosition[v, Max[v]]}]  But... why don't you sole analitically the problem? The command z /. Solve[z^2 == x^2 y - z, z]  gives {1/2 (-1 - Sqrt[1 + 4 x^2 y]), 1/2 (-1 + Sqrt[1 + 4 x^2 y])}  so it's not surprising what and where is the maximum, because %[] /. {x -> 5, y -> 5}  gives: 1/2 (-1 + Sqrt)  Too long for a comment but... For your specific setup this can be a way with a v10 function: tt1 = Table[Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5, 1}, {y, 0, 5, 1}] With[{v = z /. tt1}, With[{m = Max[v]}, {m, Most@FirstPosition[v, m]}]]  But... Why don't you solve analitically the problem? The command z /. Solve[z^2 == x^2 y - z, z]  gives {1/2 (-1 - Sqrt[1 + 4 x^2 y]), 1/2 (-1 + Sqrt[1 + 4 x^2 y])}  so it's not surprising the maximum is located at lower-right corner of the matrix, where $$x=y=5$$, and here %[] /. {x -> 5, y -> 5}  gives: 1/2 (-1 + Sqrt)  1 answered Aug 1 '14 at 12:37 unlikely 4,6401414 silver badges3939 bronze badges Too long for a comment but... For your specific setup this can be a way tt1 = Table[Solve[z^2 == x^2 y - z, z, Method -> Reduce], {x, 0, 5, 1}, {y, 0, 5, 1}] With[{v = z /. tt1}, {Max[v], Most@FirstPosition[v, Max[v]]}]  But... why don't you sole analitically the problem? The command z /. Solve[z^2 == x^2 y - z, z]  gives {1/2 (-1 - Sqrt[1 + 4 x^2 y]), 1/2 (-1 + Sqrt[1 + 4 x^2 y])}  so it's not surprising what and where is the maximum, because %[] /. {x -> 5, y -> 5}  gives: 1/2 (-1 + Sqrt)