There is however one important remark. ##Warning:
Maximize (respectively Minimize) yielding appropriate extremal values of the function can omit some arguments where it has extremal pointsvalues. This problem can be resolved using Lagrange multipliers (see e.g. How can I implement the method of Langrange multipliers to find constrained extrema?) or solving directly adequate equations.
Solve[{5#1 x^2== +First[ xMaximize[{#1, +#2}, 2x]], ==#2}, First[x]& Maximize[@@ { 5 x^2 + x + 2, -(26/5)
<= x <= 5}, x]],
-(26/5) <= x <= 5}, x]}
Here boundary arguments were omitted, but solving this system we can get all solutions
Union @
Solve[{#1 x^4== -First[ 3Maximize[{#1, x^2#2}, -x]], 1#2}, ==x]& First[Maximize[@@ { x^4 - 3 x^2 - 1,
-Sqrt[3] <= x <= Sqrt[3]}, x]],
-Sqrt[3] <= x <= Sqrt[3]}, x]
Similar problems we can encounter in more dimensional spaces and higher order equations.
There is however one important remark. Maximize (respectively Minimize) can omit some extremal points.
Solve[{5 x^2 + x + 2 == First[ Maximize[{5 x^2 + x + 2, -(26/5) <= x <= 5}, x]],
-(26/5) <= x <= 5}, x]
Union @ Solve[{ x^4 - 3 x^2 - 1 == First[Maximize[{x^4 - 3 x^2 - 1,
-Sqrt[3] <= x <= Sqrt[3]}, x]],
-Sqrt[3] <= x <= Sqrt[3]}, x]
##Warning:
Maximize (respectively Minimize) yielding appropriate extremal values of the function can omit some arguments where it has extremal values. This problem can be resolved using Lagrange multipliers (see e.g. How can I implement the method of Langrange multipliers to find constrained extrema?) or solving directly adequate equations.
Solve[{#1 == First[ Maximize[{#1, #2}, x]], #2}, x]& @@ { 5 x^2 + x + 2,
-(26/5) <= x <= 5 }
Here boundary arguments were omitted, but solving this system we can get all solutions
Union @
Solve[{#1 == First[ Maximize[{#1, #2}, x]], #2}, x]& @@ { x^4 - 3 x^2 - 1,
-Sqrt[3] <= x <= Sqrt[3]}
Similar problems we can encounter in more dimensional spaces and higher order equations.