@nm 

You can choose symbolic values for x and y, and after that give them special values in order to simplify the result (optional).

Facxy:=proc(F::algebraic, X::name=anything, Y::name=anything)
  F=1/eval(F,[X,Y]).eval(F,Y).eval(F,X), is(F*eval(F,[X,Y])=eval(F,X)*eval(F,Y))
end:

(I have used Rouben's remark to simplify the procedure.)

 Facxy(cos(x + y + 1) + sin(x - 1)*sin(y + 2),   x=a, y=b);

Facxy(1/(x-1)*y,x=a,y=b);

I'd recommend to try a different programming and presentation style.
For example, for the curve only with animation:

@asa12 

Sorry, I cannot understand the problem. What eigenvector?   sys2 and sys3 are complex-valued so Minimize is out of the question.

@ernilesh80 

1. Yes, only the gradients of the active constrains must be computed. The objective function may be non-linear but the  conditions of the KT theorem must be verified.

2. Using DirectSearch you can find directly the global max (in a domain that makes sense, where the function is real valued).

with(DirectSearch):
a1:= 5:a2:= 10:W:= 1000:
f:= (8.680555553*((1.28*(.8*W1^.7*E1^.3-15.0*(-.4+T)^2*W1^.7-.1800000000*E1-.6199999998*W1))*sqrt(-0.5333333334e-1*E1^.3+0.5333333334e-1*W1^.3+1.000000000*(-.4+T)^2)+(1.20*(.8*W2^.6*E2^.4-24.0*(-.4+T)^2*W2^.6-.2400000000*E2-.5600000000*W2))*sqrt(-0.3333333334e-1*E2^.4+0.3333333334e-1*W2^.4+1.000000000*(-.4+T)^2)+7.0656*W1^.7*E1^.3+5.1840*W2^.6*E2^.4+(7.200*(.1706666667-1.28*T+.4*T*(4*T+40)-20*T^2))*W1^.7+(8.640*(.2133333333-1.60*T+.4*T*(5*T+40)-20*T^2))*W2^.6-3.36384*E1-4.85376*W1-3.91680*E2-2.99520*W2-(.6816*(54*W1+4*E1+105*W2+5*E2))*T-3.4560-(.624*(-4*E1-54*W1-5*E2-105*W2))*T))/T:
g:= W1*a1+W2*a2 - W <=0:
dom:=E1 = 80..100, E2 = 4..6, T = 0.1 .. 1,  W1 = 100 .. 200, W2 = 10 .. 20:
GlobalOptima( f, {g, dom},maximize);

[19181.2056127683, [E1 = 92.9095671121395, E2 = 5.08993012696828, T = .161751877713913, W1 = 172.549681895113, W2 = 13.7251590524430], 4971]

@tomleslie 

1. When ||Y - X.b|| is minimized for b and X is invertible, the minimum is 0 and is obtained for b = X^(-1).Y

You can check this in the worksheet adding
ans3 := simplify(X^(-1).Y);

and you will see that ans1 = ans2 = ans3.

(A symbolic square matrix is treated as invertible).

2.  Transpose(Y-X.b).K.((Y-X.b)   can be used only if K is positive (semi)definite.

@tomleslie 

With these assumptions X is invertible (genericity applies), so b = X^(-1).Y

You should clarify first the mathematical aspects.

1. b as a solution of the minimization problem is not unique in general.
You probably want the pseudo-inverse (Moore-Penrose). See ?MatrixInverse

2. The generalization using S:=Transpose(X.b-Y).MatrixInverse(C).(X.b-Y)
is strange because S could be unbounded (inf S  could be - infinity).

@nm 

The new ode is solved via the Bernoulli method (use infolevel). Forcing "separable" ==>

dsolve(diff(y(x), x) = 3*x^2*y(x)^2 , y(x), ['separable']);

@Adam Ledger 

Sorry but the term is a standard one. See https://en.wikipedia.org/wiki/Pathological_(mathematics)

@Adam Ledger 

@rlopez 

It depends of course on how the residuals are computed in the inplicit approach.
But the space is finite dimensional and so all the norms are equivalent (for linear models).
The main problem is here the model which I think is not adequate; even y = ax+b is much better.

@rlopez

No need for implicit functions here because the fitting curve can be inversed x = (y - A*y^3)/B, or equivalently  x = a*y - b*y^3.

The problem is that this curve passes thru (0,0) which is far from the provided data, so the "model" seems to be unsuitable.

@rlopez 

@Rouben Rostamian  

But actually Maple is wrong:  sol[1] (the red one) is not a solution! ( dy/dx should be >=0).

Edit. The solution of the Cauchy problem y(0)=a is unique (x >= 0)  for a > 0.
For a=0  there are two solutions: y = 0 and y = x^3 / 9 (Maple finds only y=0).

@quo