@srikantha087 I am not sure whether using columns, like a histogram, is your specific goal. Perhaps it is.

But if you wanted to render a surface instead, note that the surfdata command allows you to speciy the axes' ranges through very simple options. Eg,

plots:-surfdata(M, 0..0.6, 0..0.6,
                colorscheme = ["Blue", "Green", "Yellow", "Red"],
                axes=normal, labels=[x,y,"f(x,y)"], orientation=[-125,70]);

That is easier than using maptrixplot (without heights=histogram) and manually creating tickmarks formulaically.

@Felipe_123 Yes, that is both simple and terse.

If you like saving keystrokes then even in old Maple 17.02 you could use the elementwise syntax,

(expand@`*`@op)~(a1);

@lekcha The (conditional check) inquality comparison of fs and tol requires that a value of type numeric is used for fs.

So,
  fs := evalf(f(x0));
suffices.

lekcha_1.mw

The result you claim for Matlab contains floats, and the result obtained in Maple does not, so they are clearly not the same expression.

@tomleslie My point is that the issue -- including the flaw you mention -- was already mentioned, and discussed in some detail, in at least two separate previous Question threads. It doesn't serve well to have yet another duplicate Question thread on the topic.

There's another closely related duplicated topic which also has at least its own three additional Question threads started by the OP, in the past few weeks. Most of it relates to pattern-matching, and the OP's continuing expectation that Maple's type system work in a fundamentally different manner.

Unfortunately the OP has (so far) not delineated the requirements of any underlying patter-matching problem class, to the extent that a robust or comprehensive approach can be easily formulated.

@lekcha I'm out of town for a few days, but if nobody else has answered then I'll get to this when I return.

As Joe points out in his answer, what's really going on in the OP's example relates to use of type `*`, and the fact the braces {..} denote type disjunction in the manner of Or(..). The OP has mistaken the problem with this example as being about type identical. And followup comments and replies below show that the goal is actually the same topic as several of the OP's recent Question threads.

These very same ordering issues with type `&*`, and the distinctions between it and and type `*`, were already covered in at two previous Question threads made by the same person within the past few weeks.

It was previously explained to the OP that this kind of use of type '`*`'({...}) does not test a specific number of multiplicands in a product (or guarantee each is present!). See the OP's Question of July 8th, What exactly is the difference between `&` and `&*`?.

And it was explained that type `&*` can guarantee existence and number of the specified multiplicands,  but not order. And one alternative structured check against a product was suggested, which could allow unspecified order through use of andmap and membertype. (Other alternative but related queries and predicates are also possible.) See OP's Question of July 1st, how to make type selection orderless.

It's not productive to have this topic of discussion duplicated and spread across multiple Question threads.

@tomleslie This is a correct answer.

@mmcdara This a relatively longwinded and inefficient way to accomplish a simple task.

I'm sorry, but it should be mentioned, because users should not be encouraged to utilize set results (eg. from fsolve or solve, etc) in this way.

@AHSAN You don't seem to understand. Your explanation is incomplete.

What do the colors and gridlines mean, in your latest image?

How do the few fixed pairs of values for two parameters beta and lambda figure into anything like a image with a continuous gradation of color?

There is no point in merely repeating the same inadequate explanation and image. That's not very respectful. You have to explain in detail the precise connection between your mutivariate expression and the plot you want.

I am finished here. 

@AHSAN I am waiting for you to provide a proper description of your goal.

As far as I can see your latest description and image, have little (if anything) to do with your original Question and  attachment (which needed only a few corrections).

I think that guessing what exactly you might want is unproductive.

@J F Ogilvie For what magnitude of relative accuracy are you looking?

@AHSAN Your question asked about a 3D plot, with some like a legend inside some blank space (eg. within the axes).

This image you just added above is not like what you asked, at all.

You should state clearly and precisely what you want  when you first ask your Questions.

@tomleslie The OP's equation was given as,

   64/19683*t^9 - 16/729*t^7 = (-t + sqrt(t^2 - 4))/2

which is equivalent to your expr2 (though you happen to have the radical isolated on the rhs). You squared both sides of your expr2, and expanded, to obtain expr3. But not all real roots of expr3 are actually roots of expr2. In particular the approximate root 2.1213203436 of your expr3 (or your expr4) is not a root of your expr2 or of the OP's equation.

The only real root of the OP's equation that is greater than 2 is given by the approximation 2.5579901741.

Also, if only real roots are wanted then there's little point to supplying the complex option to fsolve, only to have to further select the purely real roots from those.

I understand the appeal of creating your expr4, since as a univariate polynomial fsolve will find all roots for it, automatically. But the roots of your expr4 would need to be each checked for validity as roots of your expr2.

If the OP's equation is assigned to eqn then the fact that it has only a single real root greater than 2 can also be gauged by,

fsolve(eqn, t = 2 .. infinity);
fsolve(eqn, t = 2 .. infinity, avoid = {t = %});

@grheppler The first link in the Answer above starts off with this bullet-point for OSX,

I put the emphasis on that word Preferences, to distinguish it from the word Options.