Maximum area of triangle and quadrilateral - given perimeter

In summary, amongst triangles with base $a$ and perimeter $a+b$, the maximum area is obtained when the other two sides have equal length $b/2$. This can be shown using Heron's formula and the AM-GM inequality. To extend this result to quadrilaterals of given perimeter, we can divide the quadrilateral into two triangles using the diagonals, and see that the sides of the quadrilateral must be equal for maximum area. Therefore, the square has maximum area amongst quadrilaterals with given perimeter. Another approach is to consider the locus of vertex A in an ellipse with foci B and C, and see that the maximum area occurs when A is directly over the midpoint of BC, resulting in an isos
  • #1
Saitama
4,243
93
Problem:
Let $0<a<b$

i)Show that amongst the triangles with base $a$ and perimeter $a+b$, the maximum area is obtained when the other two sides have equal length $b/2$.

ii)Using the result (i) or otherwise show that amongst the quadrilateral of given perimeter, the square has maximum area.

Attempt:
Let $BC=a$ and $AB+AC=b$. From Heron's formula,
$$\Delta =\sqrt{s(s-BC)(s-AB)(s-AC)}$$
Since $s$ is given, we maximise $\sqrt{(s-BC)(s-AB)(s-AC)}$. From AM-GM inequality,
$$\sqrt{(s-BC)(s-AB)(s-AC)}\leq \frac{s^{3/2}}{3\sqrt{3}}$$
For the equality to hold, $s-BC=s-AB=s-AC$ which shows $AB=AC=b/2$. This completes the first part of problem.

How do I start with ii)? :confused:

Any help is appreciated. Thanks!
 
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  • #2
If you divide the quadrilateral into two triangles in two ways, using the diagonals, you see (from (i)) that to have maximum area it must have all four sides equal. So it is a rhombus, and it's not hard to see that the area of a rhombus with given sides is maximised when it is a square.
 
  • #3
Opalg said:
If you divide the quadrilateral into two triangles in two ways, using the diagonals, you see (from (i)) that to have maximum area it must have all four sides equal. So it is a rhombus, and it's not hard to see that the area of a rhombus with given sides is maximised when it is a square.

I don't see it. Where did you use the fact the perimeter of quadrilateral is fixed? :confused:
 
  • #4
Pranav said:
I don't see it. Where did you use the fact the perimeter of quadrilateral is fixed? :confused:
Call the quadrilateral $ABCD$, and divide it into two triangles $ABC$ and $ACD$ by the diagonal $AC$. If the lengths of $AB$ and $BC$ are not equal then (by (i)) you can increase the area of $ABC$ by making them equal (while keeping their sum constant). That will increase the area of the quadrilateral while keeping its perimeter fixed. So for maximum area $AB$ and $BC$ must have equal length. Similarly, using the other diagonal $BD$ to split the quadrilateral into two triangles, you see that $BC$ and $CD$ must have equal length.
 
  • #5
Opalg said:
Call the quadrilateral $ABCD$, and divide it into two triangles $ABC$ and $ACD$ by the diagonal $AC$. If the lengths of $AB$ and $BC$ are not equal then (by (i)) you can increase the area of $ABC$ by making them equal (while keeping their sum constant). That will increase the area of the quadrilateral while keeping its perimeter fixed. So for maximum area $AB$ and $BC$ must have equal length. Similarly, using the other diagonal $BD$ to split the quadrilateral into two triangles, you see that $BC$ and $CD$ must have equal length.

Thank you Opalg once again! :)

I hope you don't mind me posting so many problems currently. I appreciate the help I have received so far. :eek:
 
  • #6
Hello, Pranav!

Another approach . . .

Let $0<a<b$
i) Show that amongst the triangles with base $a$
and perimeter $a+b$, the maximum area is obtained
when the other two sides have equal length $b/2$.

We have \(\displaystyle \Delta ABC\) with \(\displaystyle BC = a\) and \(\displaystyle AB+AC = b.\)
Code:
                . . .
          .               .  A
        .                   o
       .                *  * .
                    *     *
      .         *        *    .
      .     o * * * * * o     .
      .     B     a     C     .
     
       .                     .
        .                   .
          .               .
                . . .
The locus of vertex [tex]A[/tex] is an ellipse with foci [tex]B,\,C.[/tex]

The maximum area occurs with maximum height:
. . when [tex]A[/tex] is directly over the midpoint of [tex]BC.[/tex]

Therefore, maximum area when [tex]\Delta ABC[/tex] is isosceles:
. . the equal sides are [tex]\tfrac{b}{2}.[/tex]
 
  • #7
soroban said:
Hello, Pranav!

Another approach . . .


We have \(\displaystyle \Delta ABC\) with \(\displaystyle BC = a\) and \(\displaystyle AB+AC = b.\)
Code:
                . . .
          .               .  A
        .                   o
       .                *  * .
                    *     *
      .         *        *    .
      .     o * * * * * o     .
      .     B     a     C     .
     
       .                     .
        .                   .
          .               .
                . . .
The locus of vertex [tex]A[/tex] is an ellipse with foci [tex]B,\,C.[/tex]

The maximum area occurs with maximum height:
. . when [tex]A[/tex] is directly over the midpoint of [tex]BC.[/tex]

Therefore, maximum area when [tex]\Delta ABC[/tex] is isosceles:
. . the equal sides are [tex]\tfrac{b}{2}.[/tex]

This is a great approach, thank you soroban! (Clapping)
 

FAQ: Maximum area of triangle and quadrilateral - given perimeter

1. What is the maximum area of a triangle with a given perimeter?

The maximum area of a triangle with a given perimeter is achieved when the triangle is equilateral. This means that all three sides are equal in length and the angles are all 60 degrees. This results in a formula for the maximum area of a triangle given a perimeter:
Area = (perimeter/3)^2 * (sqrt(3)/4)

2. How do you find the maximum area of a quadrilateral with a given perimeter?

The maximum area of a quadrilateral with a given perimeter can be found by creating a rectangle with the same perimeter. This is because a rectangle has the largest area among all quadrilaterals with the same perimeter. The formula for the maximum area of a rectangle given a perimeter is:
Area = (perimeter/4)^2

3. Can the maximum area of a triangle or quadrilateral ever be exceeded with a given perimeter?

No, it is not possible to exceed the maximum area of a triangle or quadrilateral with a given perimeter. This is because the shapes with the maximum area, an equilateral triangle and a rectangle, are already the most efficient in terms of maximizing area with a given perimeter.

4. How does the shape of a triangle or quadrilateral affect its maximum area with a given perimeter?

The shape of a triangle or quadrilateral can greatly affect its maximum area with a given perimeter. As mentioned before, an equilateral triangle and a rectangle have the largest areas among all triangles and quadrilaterals with the same perimeter. On the other hand, shapes with unequal sides and angles, such as scalene triangles and irregular quadrilaterals, will have smaller maximum areas.

5. Can the maximum area of a triangle or quadrilateral change if the perimeter is varied?

Yes, the maximum area of a triangle or quadrilateral can change if the perimeter is varied. As the perimeter increases, the maximum area of a triangle or quadrilateral will also increase. This is because there is more space to distribute the perimeter, resulting in larger areas. Similarly, as the perimeter decreases, the maximum area will also decrease.

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