funx=qLog[a1x+a1y-14]+r(100-c
2
a1x
-ix)+9/10(qLog[a2x+a2y-14]+r(100-(c
2
a2x
)/(1+bix)))​​funy=qLog[a1x+a1y-14]+r(100-c
2
a1y
-iy)+9/10qLog[a2x+a2y-14]+r100-(c
2
a2y
)(1+biy)
​​D[qLog[a1x+a1y-14]+r(100-c
2
a1x
-ix)+910(qLog[a2x+a2y-14]+r(100-(c
2
a2x
)/(1+bix))),a2x]
Out[]=
9
10
q
-14+a2x+a2y
-
2a2xcr
1+bix
In[]:=
Solve
9
10
q
-14+a2x+a2y
-
2a2xcr
1+bix
0,a2x
Out[]=
a2x
252cr-18a2ycr-
72c(9q+9bixq)r+
2
(252cr-18a2ycr)
36cr
,a2x
252cr-18a2ycr+
72c(9q+9bixq)r+
2
(252cr-18a2ycr)
36cr

In[]:=
DqLog[a1x+a1y-14]+r(100-c
2
a1y
-iy)+910qLog[a2x+a2y-14]+r100-(c
2
a2y
)(1+biy),a2y
Out[]=
9
10
q
-14+a2x+a2y
-
2a2ycr
1+biy
In[]:=
Solve
9
10
q
-14+a2x+a2y
-
2a2ycr
1+biy
0,a2y
Out[]=
a2y
252cr-18a2xcr-
72c(9q+9biyq)r+
2
(252cr-18a2xcr)
36cr
,a2y
252cr-18a2xcr+
72c(9q+9biyq)r+
2
(252cr-18a2xcr)
36cr

From testing I know at this stage it doesn’t matter if I take the positive or negative root for substitution (see below, I sub in the positive root to solve for a2y and the negative root to solve for a2x)
In[]:=
Solvea2y==
252cr-18a2ycr+
72c(9q+9bixq)r+
2
(252cr-18a2ycr)
36cr
,a2y
Out[]=
a2y
1
2
7-
q+bixq+49cr
c
r
,a2y
1
2
7+
q+bixq+49cr
c
r

In[]:=
Solvea2x==
252cr-18a2xcr-
72c(9q+9biyq)r+
2
(252cr-18a2xcr)
36cr
,a2x
Out[]=
a2x
1
2
7-
q+biyq+49cr
c
r
,a2x
1
2
7+
q+biyq+49cr
c
r

Now that a2x and a2y are in terms of ix (and a bunch of other parameters), sub back into funx and funy, using the positive root
In[]:=
funx=qLog[a1x+a1y-14]+r(100-c
2
a1x
-ix)+9/10(qLog[a2x+a2y-14]+r(100-(c
2
a2x
)/(1+bix)))/.a2x
1
2
7+
q+biyq+49cr
c
r
,a2y
1
2
7+
q+bixq+49cr
c
r

Out[]=
(100-
2
a1x
c-ix)r+qLog[-14+a1x+a1y]+
9
10
r100-
c
2
7+
q+biyq+49cr
c
r
4(1+bix)
+qLog-14+
1
2
7+
q+bixq+49cr
c
r
+
1
2
7+
q+biyq+49cr
c
r

In[]:=
funy=qLog[a1x+a1y-14]+r(100-c
2
a1y
-iy)+9/10qLog[a2x+a2y-14]+r100-(c
2
a2y
)(1+biy)/.a2x
1
2
7+
q+biyq+49cr
c
r
,a2y
1
2
7+
q+bixq+49cr
c
r
​​
Out[]=
(100-
2
a1y
c-iy)r+qLog[-14+a1x+a1y]+
9
10
r100-
c
2
7+
q+bixq+49cr
c
r
4(1+biy)
+qLog-14+
1
2
7+
q+bixq+49cr
c
r
+
1
2
7+
q+biyq+49cr
c
r

In[]:=
D(100-
2
a1x
c-ix)r+qLog[-14+a1x+a1y]+
9
10
r100-
c
2
7+
q+biyq+49cr
c
r
4(1+bix)
+qLog-14+
1
2
7+
q+bixq+49cr
c
r
+
1
2
7+
q+biyq+49cr
c
r
,a1x
Out[]=
q
-14+a1x+a1y
-2a1xcr
In[]:=
Solve
q
-14+a1x+a1y
-2a1xcr0,a1x
Out[]=
a1x
28cr-2a1ycr-
8cqr+
2
(28cr-2a1ycr)
4cr
,a1x
28cr-2a1ycr+
8cqr+
2
(28cr-2a1ycr)
4cr

In[]:=
D(100-
2
a1y
c-iy)r+qLog[-14+a1x+a1y]+
9
10
r100-
c
2
7+
q+bixq+49cr
c
r
4(1+biy)
+qLog-14+
1
2
7+
q+bixq+49cr
c
r
+
1
2
7+
q+biyq+49cr
c
r
,a1y
Out[]=
q
-14+a1x+a1y
-2a1ycr
In[]:=
Solve
q
-14+a1x+a1y
-2a1ycr0,a1y
Out[]=
a1y
28cr-2a1xcr-
8cqr+
2
(28cr-2a1xcr)
4cr
,a1y
28cr-2a1xcr+
8cqr+
2
(28cr-2a1xcr)
4cr

In[]:=
Solvea1x==
28cr-2a1xcr+
8cqr+
2
(28cr-2a1xcr)
4cr
,a1x
Out[]=
a1x
1
2
7-
q+49cr
c
r
,a1x
1
2
7+
q+49cr
c
r

In[]:=
Solvea1y==
28cr-2a1ycr+
8cqr+
2
(28cr-2a1ycr)
4cr
,a1y
Out[]=
a1y
1
2
7-
q+49cr
c
r
,a1y
1
2
7+
q+49cr
c
r

Sub a1x and a1y back into funx
In[]:=
funx=(100-
2
a1x
c-ix)r+qLog[-14+a1x+a1y]+
9
10
r100-
c
2
7+
q+biyq+49cr
c
r
4(1+bix)
+qLog-14+
1
2
7+
q+bixq+49cr
c
r
+
1
2
7+
q+biyq+49cr
c
r
/.a1x
1
2
7+
q+49cr
c
r
,a1y
1
2
7+
q+49cr
c
r

Out[]=
r100-ix-
1
4
c
2
7+
q+49cr
c
r
+qLog-7+
q+49cr
c
r
+
9
10
r100-
c
2
7+
q+biyq+49cr
c
r
4(1+bix)
+qLog-14+
1
2
7+
q+bixq+49cr
c
r
+
1
2
7+
q+biyq+49cr
c
r

In[]:=
Simplifyr100-ix-
1
4
c
2
7+
q+49cr
c
r
+qLog-7+
q+49cr
c
r
+
9
10
r100-
c
2
7+
q+biyq+49cr
c
r
4(1+bix)
+qLog-14+
1
2
7+
q+bixq+49cr
c
r
+
1
2
7+
q+biyq+49cr
c
r
