# How can't we find the maximum value of this?

Maximize[{ω, -(π/6) + (
k π)/ω <= π, -(π/
6) + ((k + 1) π)/ω >= (4 π)/3, ω > 0,
k ∈ Integers}, {k, ω}] // FullSimplify

In[446]:=
Maximize[{ω, -(π/6) + (
k π)/ω <= π, -(π/
6) + ((k + 1) π)/ω >= (4 π)/3, ω > 0,
k ∈ Integers}, {k, ω}] // FullSimplify

During evaluation of In[446]:= Maximize::mixdom: Exact optimization with mixed real and integer variables is not yet implemented.

Out[446]= Maximize[{ω, (6 k)/ω <= 7, (
2 (1 + k))/ω >= 3, ω > 0,
k ∈ Integers}, {k, ω}]
$$$$


## 2 Answers

Take the integer restriction off k and you get a good hint.

Maximize[{\[Omega], -(Pi/6) + (k*Pi)/\[Omega] <= Pi, -(Pi/6) + ((k + 1)*Pi)/\[Omega] >= (4*Pi)/3, \[Omega] > 0}, {\[Omega]}]

(*{Piecewise[{{(2*(k + 1))/3, Inequality[-1, Less, k, LessEqual, 7/2]}},-Infinity], {\[Omega] -> Piecewise[{{(2*(k + 1))/3, Inequality[-1, Less, k, LessEqual,
7/2]}}, Indeterminate]}}*)


From this you know that -1 < k <= 7/2

So make a table

Table[{k, Maximize[{\[Omega], -(\[Pi]/6) + (k \[Pi])/\[Omega] <= \[Pi], -(\[Pi]/6) + ((k + 1) \[Pi])/\[Omega] >= (4 \[Pi])/3, \[Omega] > 0},{\[Omega]}]}, {k, 0, 5}]

{{0, {2/3, {\[Omega] -> 2/3}}}, {1, {4/3, {\[Omega] -> 4/3}}}, {2, {2,{\[Omega] -> 2}}}, {3, {8/3, {\[Omega] -> 8/3}}}, {4, {-\[Infinity],{\[Omega] -> Indeterminate}}}, {5, {-\[Infinity], {\[Omega] -> Indeterminate}}}}


And now we know the maximum occurs at k = 3 if k must be an integer. Otherwise it occurs at k = 7/2.

modified

NMaximize

NMaximize[{\[Omega], -(\[Pi]/6) + (k \[Pi])/\[Omega] <= \[Pi], -(\[Pi]/6) + ((k + 1) \[Pi])/\[Omega] >= (4 \[Pi])/3, \[Omega] > 0, k \[Element] Integers}, {k, \[Omega]}]
(*{2., {k -> 2, \[Omega] -> 2.}}*)


gives wrong result!

Reduce helps to find the maximum:

cond=Reduce[{-(\[Pi]/6) + (k \[Pi])/\[Omega] <= \[Pi], -(\[Pi]/6) + ((k + 1) \[Pi])/\[Omega] >= (4 \[Pi])/3,Element[k, PositiveIntegers], \[Omega] > 0}, \[Omega]]
(*(k == 1 && 6/7 <= \[Omega] <= 4/3) || (k == 2 && 12/7 <= \[Omega] <= 2) || (k == 3 && 18/7 <= \[Omega] <= 8/3)*)

Maximize[{\[Omega], cond[[All, 2]]}, \[Omega]]
(*{8/3, {\[Omega] -> 8/3}}*)

• This maximum value has an accurate value of 8/3 Jun 8, 2023 at 7:06
• @csn899 Don' tknow why NMaximize`gives wrong result. See my modified answer. Jun 8, 2023 at 7:49