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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,237

Hi;

Find the maximum area of the rectangle formed by the points(0,0),(a,0),(0,b) and(a,b) where (a,b) lies on the curve y = (2-x^3)^(1/3) for 0 < x < 2^(1/3).

This post appears in another thread and is well answered there but what do you do if you are not as bright as those people. Faint? Say that you had a solution and the dog ate it up. Or should you exclaim in Latin that you have a marvelous answer but your briefcase is too small to contain it?

All of the above are good but better is to use geogebra. Here is how.

1) Create point (0,0) and call it A.

2) In the input bar enter f(x) = (2 - x^3)^(1/3).

3) Make a slider called a and range it from 0 to 2^(1/3).

4) Create point (a, f(a)) and call it D.

5) Create point (x(D), 0) and call it B.

6) Create point (0, y(D)) and call it C.

7) Create polygon ABCD. You can see that moving the slider creates a dynamic rectangle that always has point D on f(x) as is required.

8) Move the slider until you get the largets rectangle by checking poly1 in the algebra pane.

9) Adjust the slider increment until you are convinced that the maximum area is 1 or very close to it.

You might see something like my drawing below.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,237

Hi;

Mine is a geogebra solution ( this thread is where it belongs ). It is useful when an analytical method is unknown and an answer is needed.

The whole thread is like that, it is devoted to computer algebra systems and the results they can get.

I do not necessarily agree that your solution is easy. I suspect everyone used calculus on this one and missed it entirely. What I do think is that the geogebra solution was even easier.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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