Matrix using For Loop

Question

I am trying to create matrix, where every element is generated from For Loop. Just for simplicity,let's consider the following mathematical problem: Find maximum value of the function $x^2+y^2$ over the domains: $[0,.01],[0,.02],..,[0,1]$. In other words it is needed to create the following matrix: \begin{bmatrix}max:x^2+y^2 over [0,0.01] & max:x^2+y^2 over [0,0.02] & .. & max:x^2+y^2 over [0,0.1] \\max:x^2+y^2 over [0,0.11] & max:x^2+y^2 over [0,0.12] & .. & max:x^2+y^2 over [0,0.2]\\ max:x^2+y^2 over [0,0.91] & max:x^2+y^2 over [0,0.92] & .. & max:x^2+y^2 over [0,1]\end{bmatrix} So to automatically generate the matrix consisting from the extreme values of the function over different domains I use the following code:

 m=Table[1^i*1^j*For[k=.01,k<=1,k+=.01,
For[z=.01,z<=1,z+=.01,
Print[Maximize[{x^2+y^2,0<=x<=k,0<=y<=z,{x,y}],{i,10},{j,10}]]] 

It turns out that it doesn't work! How can I fix this problem. $$P.S.$$ Mathematically this problem is straightforward, I am interested to model this problem in Wolfram Mathematica.

Answer 1

I am not guaranteeing that this is the fastest method, but its a fairly direct translation of what you did:

Array[
 MaxValue[
  x^2+y^2,
   {x,y} ∈ Rectangle[{0,0},{0.01+0.1#1+0.01#2,0.01+0.1#1+0.01#2}]
 ]&,
  {10,10},{{0,9}}
]

Answer 2

Table[k = (i - 1)/10 + j/100;  Maximize[x^2, 0 <= x <= k, {x}][[1]], {i, 10}, {j, 10}]

Answer 3

Here is my approach.

    val = Table[{{0, 0}, {i, i}}, {i, 1/100, 1, 1/100}];
Partition[ MaxValue[x^2 + y^2, {x, y} ∈ Rectangle[First@#, Last@#]] & /@ val, 10]

Answer 4

bounds = Partition[Range[1/100, 1, 1/100], 10];
objfunc[x_, y_] := x^2 + y^2;
obj = MaxValue[{objfunc[x, y], 0 <= x <= #, 0 <= y <= #}, {x, y}]&;
Map[obj, bounds, {-1}] // MatrixForm // TeXForm

$\left( \begin{array}{cccccccccc} \frac{1}{5000} & \frac{1}{1250} & \frac{9}{5000} & \frac{2}{625} & \frac{1}{200} & \frac{9}{1250} & \frac{49}{5000} & \frac{8}{625} & \frac{81}{5000} & \frac{1}{50} \\ \frac{121}{5000} & \frac{18}{625} & \frac{169}{5000} & \frac{49}{1250} & \frac{9}{200} & \frac{32}{625} & \frac{289}{5000} & \frac{81}{1250} & \frac{361}{5000} & \frac{2}{25} \\ \frac{441}{5000} & \frac{121}{1250} & \frac{529}{5000} & \frac{72}{625} & \frac{1}{8} & \frac{169}{1250} & \frac{729}{5000} & \frac{98}{625} & \frac{841}{5000} & \frac{9}{50} \\ \frac{961}{5000} & \frac{128}{625} & \frac{1089}{5000} & \frac{289}{1250} & \frac{49}{200} & \frac{162}{625} & \frac{1369}{5000} & \frac{361}{1250} & \frac{1521}{5000} & \frac{8}{25} \\ \frac{1681}{5000} & \frac{441}{1250} & \frac{1849}{5000} & \frac{242}{625} & \frac{81}{200} & \frac{529}{1250} & \frac{2209}{5000} & \frac{288}{625} & \frac{2401}{5000} & \frac{1}{2} \\ \frac{2601}{5000} & \frac{338}{625} & \frac{2809}{5000} & \frac{729}{1250} & \frac{121}{200} & \frac{392}{625} & \frac{3249}{5000} & \frac{841}{1250} & \frac{3481}{5000} & \frac{18}{25} \\ \frac{3721}{5000} & \frac{961}{1250} & \frac{3969}{5000} & \frac{512}{625} & \frac{169}{200} & \frac{1089}{1250} & \frac{4489}{5000} & \frac{578}{625} & \frac{4761}{5000} & \frac{49}{50} \\ \frac{5041}{5000} & \frac{648}{625} & \frac{5329}{5000} & \frac{1369}{1250} & \frac{9}{8} & \frac{722}{625} & \frac{5929}{5000} & \frac{1521}{1250} & \frac{6241}{5000} & \frac{32}{25} \\ \frac{6561}{5000} & \frac{1681}{1250} & \frac{6889}{5000} & \frac{882}{625} & \frac{289}{200} & \frac{1849}{1250} & \frac{7569}{5000} & \frac{968}{625} & \frac{7921}{5000} & \frac{81}{50} \\ \frac{8281}{5000} & \frac{1058}{625} & \frac{8649}{5000} & \frac{2209}{1250} & \frac{361}{200} & \frac{1152}{625} & \frac{9409}{5000} & \frac{2401}{1250} & \frac{9801}{5000} & 2 \\ \end{array} \right)$

For the specific case in OP, x^2 + y^2 is increasing in both arguments and, thus, its maximum value is reached at the upper right corner of the constraint set (which in OP's case is a square). So we can simply square the corner coordinates and double them to get the desired result:

2 bounds ^2
% // TeXForm

same picture as above

Note: This result is the same as the one produces by the Rationalized version of @chuy's answer:

2 bounds^2 == Array[MaxValue[x^2 + y^2, {x, y} \[Element] 
    Rectangle[{0, 0}, Rationalize @ {.01+ .1 #1 + .01 #2, .01 + .1 #1 + .01 #2}]] &, 
     {10, 10}, {{0, 9}}]  

True