I have a objective function H, which follows following constrains:
p-c-a>0&&p>c>a&&0<p<1&&0<c<1&&0<a<1&&0<f[s]&&
′
f
[s]>0&&
′′
f
[s]<0

To have unique global maxima, I am assuming hessian of H to be negative definite. So, one more condition I got from here.
But when I am using Maximize[{obj, const}, {var}] I am not able to get answer. Please help!
In[]:=
H=(-a-c+p)1-
2p-f[s]
k
2
f[s](2
k
1
+
k
2
)
Out[]=
(-a-c+p)1-
2p-f[s]
k
2
f[s](2
k
1
+
k
2
)
In[]:=
D[H,{{p,s},2}]
Out[]=
-
4
f[s](2
k
1
+
k
2
)
,
2(-a-c+p)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
+
k
2
′
f
[s]
f[s](2
k
1
+
k
2
)
+
(2p-f[s]
k
2
)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
,
2(-a-c+p)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
+
k
2
′
f
[s]
f[s](2
k
1
+
k
2
)
+
(2p-f[s]
k
2
)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
,(-a-c+p)-
2
k
2
2
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
-
2(2p-f[s]
k
2
)
2
′
f
[s]
3
f[s]
(2
k
1
+
k
2
)
+
k
2
′′
f
[s]
f[s](2
k
1
+
k
2
)
+
(2p-f[s]
k
2
)
′′
f
[s]
2
f[s]
(2
k
1
+
k
2
)

For Hessian matrix [A B, C D] to be negative definite require A<0 and A D-BC>0Here as f[s], k1, k2 >0 hence
-
4
f[s](2
k
1
+
k
2
)
<0 for sureChecking AD- BC
In[]:=
Simplify-
4
f[s](2
k
1
+
k
2
)
(-a-c+p)-
2
k
2
2
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
-
2(2p-f[s]
k
2
)
2
′
f
[s]
3
f[s]
(2
k
1
+
k
2
)
+
k
2
′′
f
[s]
f[s](2
k
1
+
k
2
)
+
(2p-f[s]
k
2
)
′′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
-
2(-a-c+p)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
+
k
2
′
f
[s]
f[s](2
k
1
+
k
2
)
+
(2p-f[s]
k
2
)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
2(-a-c+p)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)
+
k
2
′
f
[s]
f[s](2
k
1
+
k
2
)
+
(2p-f[s]
k
2
)
′
f
[s]
2
f[s]
(2
k
1
+
k
2
)

Out[]=
-
4(
2
(a+c)
2
′
f
[s]
+2p(-a-c+p)f[s]
′′
f
[s])
4
f[s]
2
(2
k
1
+
k
2
)
In[]:=
Solve-
4(
2
(a+c)
2
′
f
[s]
+2p(-a-c+p)f[s]
′′
f
[s])
4
f[s]
2
(2
k
1
+
k
2
)
0,f[s]
Out[]=
f[s]
2
(a+c)
2
′
f
[s]
2(a+c-p)p
′′
f
[s]

So, for Hessian to be negative definite we have to have f[s]>
2
(a+c)
2
′
f
[s]
2(a+c-p)p
′′
f
[s]
In[]:=
MaximizeH,f[s]>
2
(a+c)
2
′
f
[s]
2(a+c-p)p
′′
f
[s]
&&p-c-a>0&&0<p<1&&0<c<1&&0<a<1&&0<f[s]&&
′
f
[s]>0&&
′′
f
[s]<0,{p,s}
Out[]=
Maximize(-a-c+p)1-
2p-f[s]
k
2
f[s](2
k
1
+
k
2
)
,f[s]>
2
(a+c)
2
′
f
[s]
2(a+c-p)p
′′
f
[s]
&&-a-c+p>0&&0<p<1&&0<c<1&&0<a<1&&0<f[s]&&
′
f
[s]>0&&
′′
f
[s]<0,{p,s}