@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 

But whose slope?

But in your approximations appears gamma itself, for which evalf is used!

I think that "solid" people parametrize and obtain the result if they want. Thank you for your thanks.

@Markiyan Hirnyk 

@Markiyan Hirnyk 

You need other tools for such problems. E.g. for the area:

f:=(1/10)*x^2-(69/790)*x+5569/255960-(3/79)*x*y-(77/4266)*y+(1/4)*y^4-(1/3)*y^3+(1/6)*y^2 - 3/1330:
evalf(Int( piecewise(f<0,1,0), x=0 .. 1, y=0..1));

       .1706012052

For a symbolic result, parametrize first. ==>

convert does not replace a space by \n; it simply inserts some \n 's.

Here is a test:

Just for fun, let's find for how many lists is the result of addList correct.
Denote by f(n) the number of "correct" lists with n elements.
Obviously such a list must have the entries in {±1, ±2,...,, ±n}.

f:=proc(n::posint)
   local v, i, P:=Iterator:-CartesianProduct([seq(1..n),seq(-n..-1)]$n);
   add(`if`(add(v) = add(v[v[i]],i=1..n),1,0), v=P) 
   end:

seq('f'(k)=f(k),k=1..6);
     f(1) = 2, f(2) = 12, f(3) = 78, f(4) = 776, f(5) = 9520, f(6) = 172932

It seems that the sequence f(n) is a serious candidate for  https://oeis.org/

t=x was just a guess.
y must be >0 if the solution is supposed to pe real (otherwise y^(25/6) is complex in general).
You will need numeric solutions, so the value for x and the initial condition must be given.
[I suspect that the ODE does not have real solutions computable by Maple].

@mehdi jafari 

@Earl 

The procedure ends because only the circles having the radius >= 1/200 are generated.
Note that the line if N > 500 then return was included just for safety, it could be eliminated.
The maths used in the procedure (and some generalizations) is presented in:
Jeffrey C. Lagarias, Colin L. Mallows and Allan R. Wilks,  Beyond the Descartes Circle Theorem,
The American Mathematical Monthly Vol. 109, No. 4 (2002), 338-361.

@mehdi jafari 

When x is in -10000 .. 10000, the oval region inside -6 .. 6  is too small to be seen or even detected by the implicitplot type procedure which is actually used. Note that if you use the suggestion in my answer then the feasible region can be obtained symbolically and explicit, so the plot will be very accurate.

@Ramakrishnan 

eq:=1/(x*y^(2/3))*8.620689655172415*10^(-16)*(-3.11*10^23*x^2*y^(7/6)-3.92*10^19*y^(25/6)+2.14545039999999*10^29*(0.0108*exp(-45.07/y)+exp(-19.98/y^(1/3)-0.00935317203476387*y^2)))/(x+0.015*y^(1.2)):
plots:-inequal(eq>0, x=-10..10, y=0..100, color=yellow);

@mehdi jafari 

You should know better (being your inequality) but usually an inequality has real sub-expressions and e.g. it has y^(7/6).
By the way, usually the domain must by specified (at least when it's not standard).

@ernilesh80 

Once you have a point where the Hessian is negatively defined, you can find a neighbourhood where it is still so.
You can use the minors of the Hessian and evalr.

An example of using evalr:

f:=(x,y) -> sin(x^2-y)*sqrt(x+y)+1.8:
f(1.,2.);
                          0.342529501
evalr(f( INTERVAL(0.9 .. 1.1), INTERVAL(1.9 ... 2.1) ));
              INTERVAL(0.081205985 .. 0.734869569)

So f is > 0 not only at the point (1,2)  but also for x in  0.9 .. 1.1,  y in 1.9 .. 2.1

@ernilesh80 

In my answer I have shown that your function is not concave (for your first set of parameters) and I suggested a way to show ("probabilistically") the concavity for other sets of parameters.

If your function is not concave and want to prove that still a local max is a global max then there are other mathematical notions such as pseudoconcavity. But this is already not Maple related, it is convex analysis.