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The perimeter of the parallelogram ABCD is 14, therefore 14=2(AB+AD) so AB+AD=7. I know that the sizes of AB, BC, CD and AD are natural numbers. How can I find the maximum area of the parallelogram? It's been long since I've done geometry, would really appreciate some help.

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    $\begingroup$ The area is maximized when the angles are right angles, and the sides are as close to equality as possible. But in this case, one can just experiment, there are very few possibilities. $\endgroup$ – André Nicolas Jun 14 '14 at 16:58
  • $\begingroup$ So it's only worth experimenting 2 cases: 1) AD=BC=4, AB=CD=3 and 2) AD=BC=3, AB=CD=4. But for these 2 cases, how do you find the size of the height? $\endgroup$ – Alex Dima Jun 14 '14 at 17:09
  • $\begingroup$ Well, three. Sides $3,4$, $2,5$, $1,6$. And $3,4$ wins. $\endgroup$ – André Nicolas Jun 14 '14 at 17:11
  • $\begingroup$ And how do you calculate the area after you figured it out that 3,4 wins? You still need to find the height, because area A=base*height $\endgroup$ – Alex Dima Jun 14 '14 at 17:17
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I will imagine that the labels $A,B,C,D$ run, as usual in mathematics, counterclockwise.

Divide the parallelogram into two triangles by cutting along the diagonal $BD$.

Now look at one of these triangles, say $\triangle ABD$. Imagine that there is a hinge at $A$. It is I think fairly easy to see that the triangle will have maximum height (with respect to base $AB$) when the angle at $A$ is a right angle.

So whatever the sides of the parallelogram are, the biggest area is achieved when the parallelogram is a rectangle.

The sum of $AB$ and $AD$ is $7$. Thus the possibilities for the legs $AB$, $AD$ are $4,3$, $5,2$, $6,1$ (and of course the same numbers reversed).

So our three candidates are rectangles of sides $4,3$, $5,2$, and $6,1$. It is clear that the rectangle with sides $4$ and $3$ wins.

Remarks: $1.$ If we had the freedom to choose non-integers, the maximum would be reached when each of $AB$ and $AD$ is $3.5$. In general, the parallelogram with given perimeter and maximum area is a square. Because of the restriction to integers, the best is when the sides are as nearly equal as possible.

$2.$ If you want a more calculational way to find that it is best to make $\angle A=90^\circ$, you can use the fact that the area of $\triangle ABD$ is $(1/2)(AB)(AD)\sin A$. But I think it is geometrically clear that we get maximum height with respect to base $AB$ when the angle is a right angle.

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  • $\begingroup$ I got that the best case is 3 and 4. But now I don't get it how can I calculate the actual area in this case. I still don't know the height of the parallelogram, and the area is A=base*height. So if BC=4, I need to calculate AE, the perpendicular from A to BC to get the actual area. $\endgroup$ – Alex Dima Jun 14 '14 at 17:25
  • $\begingroup$ The base is $4$ and the height is $3$, since for maximum area we have a right angle at $A$. So the maximum area of the parallelogram is $12$. $\endgroup$ – André Nicolas Jun 14 '14 at 17:28
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The geometric shape which occupies the largest surface given a fixed length is a circle, and the one occupying the largest volume given a fixed surface is a sphere. Now, if roundness is to be excluded, we're left with regular polygons, since the circle itself can be viewed as a regular polygon, with an infinite number of sides, each of them the size of a single point. Now, since a parallelogram is four sided, and four-sided regular polygons are squares, a preliminary answer would be a square of side length $\dfrac{14}4=\dfrac72=3.5$. But we are given the extra condition that all sides be natural numbers. So the final answer becomes a rectangle with sides $3$ and $4$. Since rectangles are parallelograms, we're done.

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