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.
Find the maximum area of a parallelogram given that you know the perimeter
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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.
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.