Items tagged with maximize


Let be the number z so that |z+3-2*I| + |z-3-8*I| = 6*sqrt(2). Find min and max of the modulus of z. How can I find min and max of modulus of z with Maple.

Thank for your help!

i have an optimization problem, i want to maximize an expression using assumption, what should i do?




M1 := Matrix(1, 4, {(1, 1) = p^(1/2), (1, 2) = 0, (1, 3) = 0, (1, 4) = (1-p)^(1/2)})



M2 := Matrix(4, 1, {(1, 1) = cos(theta[1])*cos(theta[2]), (2, 1) = exp(I*phi[1])*sin(theta[1])*cos(theta[2]), (3, 1) = exp(I*phi[2])*sin(theta[2])*cos(theta[1]), (4, 1) = exp(I*(phi[1]+phi[2]))*sin(theta[1])*sin(theta[2])})






maximize(PP) assuming 0<p ,p<1;

Error, (in assuming) when calling 'maximize'. Received: 'invalid input: `minimize/continuous` expects its 2nd argument, yFP, to be of type {name, list(name)}, but received `theta[1]` = -infinity'





When using maximize on a relatively complicated function (see attached Maple file and PDF), it runs extremely slow. No return after 3 minutes.

My hardward: i7 2.3G, 8G DDR3 MEM, 500G SSD.

Maybe someone is interested to try the Maple code if your workstation is more powerful? :)


The commands below are from a response by Carl Love to a question posed on February 27, 2016.

Variable t is not mentioned in the plot of f. Does t assume one or more particular values in the construction of the plot?

When t is given a specific value before executing the plot command, the resulting plot appears to be independent of t's value.

What is the logic behind the plot's construction?

f:= cos(2*t/m) + cos(2*(t+5)/m):
plot('maximize'(f), m= 1..10);

Consider the following expression obtained from the solve command: Note that this uses the two variable arctan function.

p:=arctan((-cos(theta)^3-(1/2)*cos(theta)^2-(1/2)*cos(theta)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)+2*cos(theta)-(1/2)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)+3/2)^(1/2), -(1/2)*cos(theta)-1/2-(1/2)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)):

#ploting the expression shows a non-zero value at theta = Pi,  however if I convert p to a function using


# then f(Pi);  gives a value of 0

#On the other hand maximize(p,theta=3*Pi/4..5*Pi/4,location); shows a non-zero value of 4*Pi/5 at theta = Pi,  which agrees with the plot of p, namely, it returns:


{[{theta = Pi}, -arctan((10-2*5^(1/2))^(1/2)/(5^(1/2)+1))+Pi]}

Is this a bug? Or what?



When I put maximize(cos(t)), everything is fine.

When I put maximize(cos(Pi)), everything is fine.

When I put maximize(cos(t*Pi)), it says invalid limiting point??? What went wrong?


Hi all,

I have a function f(x) and want to know at which x-value it attains its supremum.

Tried this, but it doesn't work (or at least hasn't been able to solve the equation in 10+ min):

M := maximize(f, x = 0 .. 1);

solve(f = M, x);

Does anyone know a way to do this?



a := 1; w0 := 1; w1 := 3; K := 10; h0 := .25; A := 3; alpha := .7; b := .6; c := .2

f := A*x^(b*alpha)*(alpha*A*x^(b*alpha)/(a+w0*h0+(x-h0)*w1))^(alpha/(1-alpha))-(alpha*A*x^(b*alpha)/(a+w0*h0+(x-h0)*w1))^(1/(1-alpha))




M := maximize(f, x = 0 .. 1)



solve(f = M, x)

Warning,  computation interrupted



plot(f, x = 0 .. 1)






I've got this huge chunk of code which leads to an optimiazation at the very last line (Bestangles:=minimize(maximize()-minimize))). This minization is taking a very long time (havent solved it yet) and I would very much like to reduce that time. As I've understood maple does optimization by differentiating and then finding all extremes and comparing. Would this mean that since I minimize and optimize within a minimization command, it differentiates a ton of times? And if this is the case, can I somehow do the differentiation beforehand, since it is the same function being differentiate all the time? Or is there some other way I can improve the code? 
Thanks alot!

Heres the full code:

I am not sure how/why, but here is the worksheet.


The function evalutes fine and can be used for sequence. But it does not seem to be working with plot or Maximize.

V is assumed to between 0 and 1.

Need some help.






  I have the following input

maximize(f, x=0..100,location);

 Maple gives me the location is x=1.25. However, how should I do to obtain this position?  If I write


a:=maximize(f, x=0..100,location);


  Seems it do not work :(


  I may try fsolve at the maximum value, but it seems to be awkward..


Thank you very much!


Hi there,

I am trying to maximize a function given a set of values to a parameter in the function. The function is an differential equation belonging to a system of two differential equations.

I have a for loop to state different values to the parameter.

Maple yields the error:

Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

When trying to maximize the function.

Supposed that I was doing something wrong in the loop, if I reproduce the contents of the loop outside, and set a value for the parameter. If I plot the solution of the ordinary differential equation, I can see where the maximum lies.

Having plot it, the Optimizamtion:-Maximize works as expected.

However, omitting the plot has a weird effect: I only get the same result depending on the bounds I set for the Maximization:

de1 := diff(A(t), t) = r*m*(1-g)*A(t)-piecewise(t < 8, r*A(t), t >= 8, (r+k)*A(t));
de2 := diff(G(t), t) = r*m*g*A(t)-l*G(t);

ics := A(0) = 25.0, G(0) = 0.;
num := dsolve({de1, de2, ics}, {A(t), G(t)}, type = numeric, output = listprocedure, parameters = [g]);

num(parameters = [g = .15]);
val := eval(G(t), num);

# odeplot(val, [t, G(t)], t = 0 .. 100);

Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

val2 := Maximize(val);

Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

val3 := Maximize(val(t), t = 0 .. 60);

  [10267.824035766165, [t = 8.25727747134303]]

val4 := Maximize(val(t), t = 0 .. 100);

[6.863211343195069e-9, [t = 59.84184367042171]]


The right answer is [10267.824035766165, [t = 8.25727747134303]]: Why do I get two different answers even if in that range there is only one relative maximum?

I ignore whether the way I am specifying the arguments for the Maximize function is correct. val is a procedure.


What am I missing?

Attached is the worksheet:




I want to solve maximize of equation,but the maximize failed to solve it,who can help me.thanks.

c[1] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)}:

c[2] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+[1/(z^2+1)]^(3/2)+1}+(1/8)*{y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1)}:

t[1] := diff(c[1], x);



t[2] := diff(c[2], y);



eliminate({t[1], t[2]}, w);

[{w = -{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}/{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}}, {{1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}-{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(x^2+2*x*z+z^2+1))^(1/2)*(2*x+2*z)/((x+z)^2+1)^2-3*(1/(x^2+1))^(1/2)*x/(x^2+1)^2}}]


w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*sqrt(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*sqrt(1/(x^2+2*x*y+y^2+1))*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*sqrt(1/(y^2+2*y*z+z^2+1))*(2*y+2*z)/((y+z)^2+1)^2-3*sqrt(1/(y^2+1))*y/(y^2+1)^2);

w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2)


sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[1]);

sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)})


subs(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[2]);



"#"Iwant to maximize the equation (5)and (6),under the conditon of x,y,z are negative or positive at the same time.





I want to find the greatest value of this expression 


with x>0, y>0 , z>0,x+y+z=3.

I tried



DirectSearch[GlobalOptima](f(x,y,z), {x>0, y>0 , z>0,x+y+z=3},maximize);

I got the output

[HFloat(infinity), [x = .591166078050740e52, y = .183647204560715e52, z = .786638021216969e52], 1249]



what is the wrong with Pi set ::: in this function ::: Warning, no iterations performed as initial point satisfies first-order conditions

Optimization[Minimize](x^2 + y^2 + 25*(sin(x)^2+sin(y)^2), x=-2*Pi .. 2*Pi , y= -2*Pi .. 2*Pi);

Warning, no iterations performed as initial point satisfies first-order conditions
[0., [x = HFloat(0.0), y = HFloat(0.0)]]

Optimization[Maximize](x^2 + y^2 + 25*(sin(x)^2+sin(y)^2), x=-2*Pi .. 2*Pi , y= -2*Pi .. 2*Pi);

Warning, no iterations performed as initial point satisfies first-order conditions
[-0., [x = HFloat(0.0), y = HFloat(0.0)]]


I got my good result when I apply it with this function :

f:= (x,y)->cos(x)*sin(y) -(x/(y^2+1));

Optimization[Maximize](f(x,y), x = -1 .. 2, y = -1 .. 1);

[0.994945017202501170,[x = HFloat(-0.6362676080636113), y = HFloat(1.0)]]

Optimization[Minimize](f(x,y), x = -1 .. 2, y = -1 .. 1);

[-2.02180678335978703,[x = HFloat(2.0), y = HFloat(0.10578346945175972)]]

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