input -->

restart; with(plots);
H := 11;
B := x; W := x;
t__E := (13100*(B*B)*H*(.23-.10)*1.2)/(1500*alpha);
A__t := 2*B*W+2*H*(B+W);
OF := alpha/A__t;
for alpha from .1 by .1 to 2 do XX[alpha] := solve({0 < x, OF < 0.6951871658e-2, t__E < 200}, x) end do

result-->

         XX[0.1] := {0.3222041888 < x, x < 1.155224359}
         XX[0.2] := {0.6354894776 < x, x < 1.633733956}
         XX[0.3] := {0.9405579047 < x, x < 2.000907284}
         XX[0.4] := {1.238023973 < x, x < 2.310448718}
         XX[0.5] := {1.528429210 < x, x < 2.583160196}
         XX[0.6] := {1.812253750 < x, x < 2.829710218}
         XX[0.7] := {2.089925662 < x, x < 3.056436363}
         XX[0.8] := {2.361828525 < x, x < 3.267467912}
         XX[0.9] := {2.628307644 < x, x < 3.465673077}
         XX[1.0] := {2.889675191 < x, x < 3.653140182}
         XX[1.1] := {3.146214498 < x, x < 3.831445747}
         XX[1.2] := {3.398183646 < x, x < 4.001814567}
         XX[1.3] := {3.645818516 < x, x < 4.165220660}
         XX[1.4] := {3.889335368 < x, x < 4.322453755}
         XX[1.5] := {4.128933055 < x, x < 4.474164703}
         XX[1.6] := {4.364794925 < x, x < 4.620897435}
         XX[1.7] := {4.597090458 < x, x < 4.763112053}
         XX[1.8] := {4.825976698 < x, x < 4.901201868}
 

question -->

anyone knows a simple way how I can plot the points of the lower boundaries and upper boundaries of x as two line graphs (without copy pasting all the values into separate lists and then using the listplot command)?

thanks

Hey!

The more "nodes, compartments, ..." I define the longer my formula (see below) gets. The formula posted below is written for a starting situation of 9 nodes (i,j,k,l,m,p,q,r,s). The problem that I have is that, besides running out of letters to chose from, the formula is getting to long and is not clear anymore (losing the overview). So I'm looking for a way to write it shorter, especially for the bold parts I assume there must be a better way.

B is a 9x9 matrix, n[0]:=1, N:=9

for i in n[0] do for j to N do for k to N do for l to N do for m to N do for o to N do for q to N do for r to N do for s to N do

if B(i, j) = 1 and B(j, k) = 1 and B(k, l) = 1 and B(l, m) = 1 and B(m, o) = 1 and B(o, q) = 1 and B(q, r) = 1 and B(r, s) = 1 and i <> j and i <> k and i <> l and i <> m and i <> o and i <> q and i <> r and i <> s and j <> k and j <> l and j <> m and j <> o and j <> q and j <> r and j <> s and k <> l and k <> m and k <> o and k <> q and k <> r and k <> s and l <> m and l <> o and l <> q and l <> r and l <> s and m <> o and m <> q and m <> r and m <> s and o <> q and o <> r and o <> s and q <> r and q <> s and r <> s

then P[loss, i, j, k, l, m, o, q, r, s] := P[loss, i, j, k, l, m, o, q, r]*P[fw, r, s]*P[suppr, i, j, k, l, m, o, q, r]*P[fo, s] else P[loss, i, j, k, l, m, o, q, r] := 0 end if; if P[loss, i, j, k, l, m, o, q, r] <> 0 then print(subs(I = i, J = j, K = k, L = l, M = m, O = o, Q = q, R = r, 'P[loss, I, J, K, L, M, O, Q, R]') = P[loss, i, j, k, l, m, o, q, r]) end if end do end do end do end do end do end do end do end do end do

I tried i<>j<>k<>l<>m<>o<>p<>q<>r<>s but that does't work

P.S.: Do you always have to specify something after ELSE using the IF loop? Actually I just don't want maple to do anything if the conditions are not met but since I then cannot print only the solutions I want (the ones for which the conditions are met) I always end up specifying something after else.

Thanks!!

hello,

this is my first post here so sorry if it's done wrong...

I'm having difficulties with the following command because maple only returns a solution if the first two boundary conditions of the "if" arguments are true. If the the first or the second condition is false then maple doesn't give a solution solution.

for i to n do if H[i] < 2.7 then if A[i, f] < 12 then if A[i, o] < 1.2 then Q[i, foo] := evalf(610*(A[i, o]*sqrt(H[i])*h[k]*A[i, T])^(1/2)) else Q[i, foo] := evalf(7.8*A[i, t]+378*A[i, o]*sqrt(H[i])) end if end if end if; print(Q[i, foo]); end do;

I've also tried with the elif command but it gives a similar problem

Anyone knows how to solve this?