How to write maximize or minimize function in J - j

Question

For example, if I want to maximize the expectation of returns function
E[r]= w1r1+w2r2 and solve the optimization value for the weight w1 and w2.

Answer 1

The only constraint that you have really given is that w1+w2=1
w1 =.0.25
(,~ -.)w1
0.25 0.75
That takes care of both w1 and w2 given the value of w1.
r1 r2 +/#:* w1 w2 calculates r1w1 + r2w2
r1 =. 5
r2 =.10
(r1,r2) (+/#:* (,-.))w1
8.75
(r1,r2) (+/#:* (,-.))0.9
5.5
(r1,r2) (+/#:* (,-.))0.01
9.95
If you really wanted to maximize you would need to add equations for the value of r1 and r2 and take those into account as well, but perhaps I don't understand your question?
Responding to the comment below: If the constraint of w1+w2=1 still is in play, then the matter just becomes summing the values in r1 and r2, then whichever is bigger should get the w value of 1 and the other will get the w value of 0
r1=.2 4 6 3 2
r2=.2.1 4 6 3 2
r3=.2 4 6 3 2.3
r1 (,-.)#:>/#:(+/#:,.) r2
0 1
r2 (,-.)#:>/#:(+/#:,.) r1
1 0
r3 (,-.)#:>/#:(+/#:,.) r2
1 0
'w1 w2'=.r3 (,-.)#:>/#:(+/#:,.) r2
w1
1
w2
0
'w1 w2'=.r1 (,-.)#:>/#:(+/#:,.) r2
w1
0
w2
1
(r1,.r2) +/#:,#:(+ . *) (0 1) NB. w1=.0 w2=.1
17.1
(r1,.r2) +/#:,#:(+ . *) (1 0) NB. w1=.1 w2=.0
17
(r1,.r2) +/#:,#:(+ . *) (0.5 0.5) NB. w1=.0.5 w2=.0.5
17.05
Based on the follow up comment below I would approach it in one of two ways. I could dig up all my linear programming texts from the 1980's and come up with the definitive mathematical solution (including degenerative cases and local maxima/ minima) or using the same technique as above but for a larger case than n=2. I'm going with the second option.
Let's look first at the r matrix which will be a set of constants. For this example I am taking a random 5 X 10 matrix with values from 1 to 10.
r=. >: ? 5 10 $ 10
r
4 4 8 1 4 3 6 9 6 2
2 6 5 4 4 7 5 10 4 6
2 4 9 10 1 1 9 8 2 7
5 6 5 4 7 9 2 6 10 6
10 3 6 2 10 2 7 10 4 2
Now the trick that I am going to use is that I want to find the column with the highest average to be multiplied by the largest value of w. Easy to do with J using (+/ % #)
(+/ % #) r
4.6 4.6 6.6 4.2 5.2 4.4 5.8 8.6 5.2 4.6
Then find the ranking of the list to be able to reorder the columns of the original r matrix. The leading 7 means that 7 { r is the largest average etc.
\:#:(+/ % #) r
7 2 6 4 8 0 1 9 5 3
I use this to in turn reorder the columns of the matrix r using {"1 since I am working columns. The result is that I have reordered the columns of r so that the column with the largest average is on the left and smallest on the right.
(\:#:(+/ % #) {"1 ]) r
9 8 6 4 6 4 4 2 3 1
10 5 5 4 4 2 6 6 7 4
8 9 9 1 2 2 4 7 1 10
6 5 2 7 10 5 6 6 9 4
10 6 7 10 4 10 3 2 2 2
Once I have that, then the next thing is to develop the w vector. Since I now have all the largest averages on the left I will just maximize the values to the left of w to be as large as possible within the noted constraints.
w=. 0.2 0.2 0.2 0.2 0.15 0.01 0.01 0.01 0.01 0.01
#w NB. w1 through w10
10
+/w NB. sum of the values in w
1
>./w NB. largest value in w
0.2
<./w NB. smallest value in w
0.01
Because the r matrix has been reordered using + . * the dot product gives values for w1r1 , w2r2 , w3r3 ... w10r10
(({"1~ \:#: (+/ % #))r) + . * w
1.8 1.6 1.2 0.8 0.9 0.04 0.04 0.02 0.03 0.01
2 1 1 0.8 0.6 0.02 0.06 0.06 0.07 0.04
1.6 1.8 1.8 0.2 0.3 0.02 0.04 0.07 0.01 0.1
1.2 1 0.4 1.4 1.5 0.05 0.06 0.06 0.09 0.04
2 1.2 1.4 2 0.6 0.1 0.03 0.02 0.02 0.02
to actually get the weight of the matrix ravel all the values then sum
+/ , (({"1~ \:#: (+/ % #))r) + . * w
31.22