Three families grow vegetables in their backyards,and agree to form a small closed economy by sharing their produce. family andrews grow artichokes, family brown grows beans and family Cuthbert grows corn.Family Andrews receives 70% of its artichokes, 30% of the beans and 30% of the corn.Family Brown receives 20% of the artichokes,60% of the beans and 10% of thr corn.Family Cuthbert receives the remainder of the vegetables.

a) Write down the exchange matrix A.

b) If production is measured in doolars ,define the variables
                                    x[1]
,
                                    x[2]
,
                                    x[3]
 of the production vector X.

c) Find all solution of the Leontief closed with matrix equation (I-A)^-1¤X=0.

d) If family Cuthbert produces $100 worth of corn, how much will familes Andrews and Brown need to produce (in dollars) in order foe the economy to be in equilibrium?

Hey everyone,

I have seen similar questions a few times but I didn't really understand why this problem occurs in my system (as well as I am ney to maple). Therefore I would appreciate every help a lot!!!

I do have the following linear differential system:

As additional conditions I use:

And solve the system by doing:

additionally I define afterwards:

The result of for example Result[1] is:

Why do I get the unknown value of _U1? In my opinion I defined all parameters for the Linear differential system.

Thanks for every help!

Markus

hello 

please say me how can i make diff and implicit diff with Set Point answer?

for example how can i write these?

(x^2+y^2=6)' => 2*x+2*y*y'=0 => y'(-2,5)=2/5

or

(y=x^3+4*x)' => y'=3*x^2+4 => y'(3)=31

thanks very much

delete this post, not maple related

apologies

How would i label the axis and change its font and size.

Also, how would i move the horizontal axis down, so that the whole diagram can be seen. (Hope that makes sence)

Finally, there are some odd points between 0 and 1 on the vertical axis, how do you get rid of these...

Digits:=20: N:=10000: M:=100: x_max:=1: r_min:=0:
r_max:=4: for n from 0 to N do r:=r_min+n/N*(r_max-r_min):
x:=evalf(x_max*rand()/10^12):for m from 0 to M do x:=x*exp(r*(1 - x)): od:
X[n]:=x: od:
with(plots):
bifpoint:={seq([r_min+j/N*(r_max-r_min),X[j]],j=0..N)}:
pitchf:=pointplot(bifpoint,symbol=point):display(pitchf);

Thanks

I am trying to solve the rate equations for A<=>B<=>C with k1,k2,k3,k4 as the rate constants.  I put in the equations into Maple but can't seem to get it to solve them.  The initial conditions for A(0) = S, B(0)=C(0)=0

Here is what I plugged into the program:

Parameters(S, k1, k2, k3, k4);

sys_ode := d*A(t)/dt = -k1*A(t)+k2*B(t), d*B(t)/dt = k1*A(t)+(-k2-k3)*B(t)+k4*C(t), d*C(t)/dt = k3*B(t)-k4*C(t);

d A(t)
------ = -k1 A(t) + k2 B(t),
dt

d B(t)
------ = k1 A(t) + (-k2 - k3) B(t) + k4 C(t),
dt

d C(t)
------ = k3 B(t) - k4 C(t)
dt

ics := A(0) = S, B(0) = 0, C(0) = 0;
A(0) = S, B(0) = 0, C(0) = 0

dsolve({ics, sys_ode});
Error, (in dsolve) found functions with same name but depending on different arguments in the given DE system: . It is required an indication of the dependent variables

I have to questions. One is the main one, about simplifying fractions.

This is equations.

jj:=L*theta*gamma[2]^2*(-1+alpha)/(delta[2]*(-alpha*gamma[1]^(-2*sigma+2)*gamma[2]^(2*sigma)-gamma[2]^2+gamma[2]^2*alpha));

kk:=expand(denom(jj)/gamma[2]^2);

ll:=op(1,kk);

mm:=simplify(ll,power);

nn:=mm-delta[2]+delta[2]*alpha;

oo:=numer(jj)/gamma[2]^2/nn;

pp:=(numer(jj)/gamma[2]^2)/(denom(jj)/gamma[2]^2);

From now, I will explain. Though as I don't know how to put a maple screen on this board I think I will have you feel some inconvenience, please forgive it.

I want to make jj more simple, dividing both of numerator and denominator of jj, with gamma[2]^2.

But, what I could at best was this. First, taking denominator of jj and deviding it with gamma[2]^2 and expanding it like kk. Second, as the first term among that expression had terms of gamma[2] on both of the numer and denom, I picked out the first term as ll in the above maple commands, and simplified it with option: power, in mm. Thirdly, I rewrote kk as in nn.

and Forthly, I rewrote jj with modification of deviding numerator and denominator with gamma[2]^2, using the result which I could get as I wrote above. oo is that. And the result of oo is what I wanted to get.

But, on the other hand, when i divided denom and numer of jj directly by gamma[2]^2, the result I can get then is not 

good one: the output of pp is not what I wanted to get.

Is there more simple way to do what I did in the above? This is a first question.

And, another is how I can post the screen of maple on this question board?

I hope you will answer my question. I thank all of you who will answer these question in advance.

Hellp

for a task about the launch of a rocket we had to make a matrix OPL3 with in the first column the time T and in the second column the height of the rocket at that time.

I must Maple find the time where the rocket is at his highest point.

I can find how high the highest point is, but didn't succeed to find the corresponding time.
Can Maple tell me at wich time (= row # o/t matrix OPL3) the rocket is at his highest point.

I used the following to find the highest point.

ymin:=min(seq(abs(OPL3(i,2)),i=64..830));

Note: the diff equitation is numerically solved, so I can't use solve(....=0) (am I right here?)

Thanks!

How  can you create a loop for Monty Hall Problem when you have 3 door (1 opening) and then 4 Doors (with 2 openings and possible 2 switches)

I have this kind of problem. When I try to solve my command:

is there any way to avoid this? in order to get the answer directly...

I encounter "insufficient  initial/boundary value" error message,  do know how to proceed from there, search with "insufficient initial value" gets no result. Any help will be appreciated.

> restart; alias(r = r(t), f = f(t)); with(plots);
r, f
> DE := diff(r, t) = 2*r+alpha*r*f, diff(f, t) = -f+alpha*r*f;
d d
--- r = 2 r + alpha r f, --- f = -f + alpha r f
dt dt
> NULL;
> params := alpha = .3;
alpha = 0.3
> initv := r(0) = 101, f(0) = 2;
> NULL;
>
> dvars := [r, t];
> chaodisplay := proc (chartname) EQ := [op(subs(params, [DE])), initv]; EQ1 := dsolve(EQ, numeric); odeplot(EQ1, dvars, t = 0 .. 300, axes = frame, numpoints = 50000, color = green, orientation = [-30, 100], title = chartname) end proc;
Warning, `EQ` is implicitly declared local to procedure `chaodisplay`
Warning, `EQ1` is implicitly declared local to procedure `chaodisplay`
>
> chaodisplay("Rabbit and Fox");
Error, (in dsolve/numeric/type_check) insufficient initial/boundary value information for procedure defined problem
>

Dear people being here,

I have a question.

When I wrote as

aa:=(delta[1]*n[1])/(n[f1]*delta[1]);

maple returned as 

aa := n[1]/n[f1]

How can I have maple express the original one, aa:=(delta[1]*n[1])/(n[f1]*delta[1]), not what returned to me?

I hope you will answer to my question.

And, thanks in advance.

taro

Please help me, here my problem:

Parsing a string works correctely like this:

ABC := 3443;

print("ABC = ", eval(parse(cat("A", "B", "C"))));

But using PROCEDURE it doesn't work:

myParse := proc ()

     local PEA;
   

      PEA := 4334;

      print("PEA = ", eval(parse(cat("P", "E", "A"))))

end proc;

myParse();

                        "PEA = ", PEA

I suppose to have "PEA = ", 4334

how to make a table of newton raphson ? 
working is below 
printf("No           root                   x[i+1]                 abs(x[i+1]-x[i])\n"); for i to n do y[i] := evalf(x[i]+(-D1(x[i])+sqrt(D1(x[i])^2-2*f(x[i])*D2(x[i])))/D2(x[i])); x[i+1] := evalf(y[i]-f(y[i])/sqrt(D1(x[i])^2-2*f(x[i])*D2(x[i]))); printf("%d         %10.10f     %10.10f       %10.10e \n", i+1, x[i+1], f(x[i+1]), abs(x[i+1]-x[i])); if abs(x[i+1]-x[i]) < tol then print("approximate solution" = x[i+1]); print("No of iterations" = i+1); break end if end do
please help me 

Hi, I need to show some how by using Maple, that n^3>(n+1)^2 where n>=3. Iam new at Maple and not sure how I can graph inequality as n goes to infinity. Any help would be appriciated. Thanks!!!