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I Have the problem with the maximize of this problem:

ClearAll[expr, max, sol, x, y];
expr = (1 + x + y)/((1 + x + y) + 1)* 10 - 
   y - (((1 + x)/((1 + x) + 1))*10);

max = Simplify[Maximize[{expr, 0 <= y <= 1.5, 0 <= x <= 0.5}, y]]

What it returned is:

Maximize[{-((y (x^2 + 2 (-3 + y) + x (4 + y)))/((2 + x) (2 + x + y))),
   0 <= y <= 1.5, 0 <= x <= 0.5}, y]

So it did not work.

I have another question with regard to extraction of value when use maximize function. After using Maximize, how do I extract the value of y and the value of the maximum function? y/. did not seem to work

My final question is regard to lower bound of a function. For example, my function is this

(1 + x + y)/((1 + x + y) + 1)* 10 - y - (((1 + x)/((1 + x) + 1))*10);

I want to create a code, such that for the above function, when x runs from 0 to 1, we find the maximum of that, expressed in terms of x. However, if the function is lower than 0, I want to return 0 and y is equal 0 in such case.

Thank you all so much!

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Clear["Global`*"]

expr = (1 + x + y)/((1 + x + y) + 1)*10 - y - (((1 + x)/((1 + x) + 1))*10);

Use exact numbers and use the constraints as Assumptions in Simplify

maxX = Simplify[
  Maximize[{expr, 0 <= y <= 3/2, 0 <= x <= 1/2}, x], {0 <= y <= 3/2, 
   0 <= x <= 1/2}]

(* {-(((-3 + y)*y)/(2 + y)), 
   {x -> Piecewise[{{1/4, y == 0}}, 0]}} *)

EDIT: Use Part to verify that the max value (maxX[[1]]) is produced by evaluating expr at the specified value of x (maxX[[2]]),

Assuming[{0 <= y <= 3/2, 0 <= x <= 1/2},
 maxX[[1]] == expr /. maxX[[2]] // Simplify]

(* True *)

maxXY = Maximize[{expr, 0 <= y <= 3/2, 0 <= x <= 1/2}, {x, y}] //
  Simplify

(* {7 - 2 Sqrt[10], {x -> 0, y -> -2 + Sqrt[10]}} *)

maxXY // N

(* {0.675445, {x -> 0., y -> 1.16228}} *)

EDIT: Again, using Part to verify,

maxXY[[1]] == expr /. maxXY[[2]]

(* True *)
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Your expression has two variables, so

max = Simplify[Maximize[{expr, 0 <= y <= 1.5, 0 <= x <= 0.5}, {x, y}]]

{0.675445, {x -> 0., y -> 1.16228}}

The maximum value is then

max[[1]]

0.675445

And the x and y values are

{x, y} /. max[[2]]

{0., 1.16228}

Update

Maybe something like this:

Your expression looks well-behaved

Plot3D[expr, {x, 0, .5}, {y, 0, 1.5}, AxesLabel -> {x, y}]

enter image description here

So we could see where the derivative with respect to y is 0

sol = Solve[D[expr, y] == 0, y]
(* {{y -> -2 - Sqrt[10] - x}, {y -> -2 + Sqrt[10] - x}} *)

Plot[y /. sol[[2]], {x, 0, .5}, AxesLabel -> {"x", "maxy"}]

enter image description here

Alternatively,

maxy[xx_] := 
 Maximize[{expr, 0 <= y <= 1.5, 0 <= x <= 0.5} /. x -> xx, y]

Plot[y /. maxy[x][[2]], {x, 0, .5}]

enter image description here

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  • $\begingroup$ that's not what I meant, I need to find maximum of function with regard to x (in economics its called value function, if you do economics) $\endgroup$ – user66418 Sep 1 at 21:39
  • $\begingroup$ @user66418 See if the answer update is more helpful $\endgroup$ – MelaGo Sep 1 at 22:08
  • $\begingroup$ Hi, thank you for your help, I asked my colleagues and they say Nsolve may miss some maximum value. Do you have anyother suggestions? Thank you so much $\endgroup$ – user66418 Sep 8 at 20:05

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