Maximum area of a Triangle

Printable View

July 4th 2009, 07:08 AM
JoAdams5000

Maximum area of a Triangle

Suppose a triangle has perimeter 2. Let
x and y be the lengths of two of its sides and A its area. Show that A^2 = (x + y − 1)(1 − x)(1 − y). Use this formula to find the maximum area of a triangle with perimeter 2.

Hint: You will need to solve a maximization problem

over a closed, bounded set D. To determine D, use the triangle inequality.

Kind of lost with where to begin. Any help and/or suggestions would be great! Thanks
July 4th 2009, 07:20 AM
malaygoel

Quote:

Originally Posted by JoAdams5000

Suppose a triangle has perimeter 2. Let
x and y be the lengths of two of its sides and A its area. Show that A^2 = (x + y − 1)(1 − x)(1 − y). Use this formula to find the maximum area of a triangle with perimeter 2.

Hint: You will need to solve a maximization problem

over a closed, bounded set D. To determine D, use the triangle inequality.

Kind of lost with where to begin. Any help and/or suggestions would be great! Thanks

Use hero's formula to determine $A^2$
July 4th 2009, 07:22 AM
malaygoel

Quote:

Originally Posted by JoAdams5000

Suppose a triangle has perimeter 2. Let
x and y be the lengths of two of its sides and A its area. Show that A^2 = (x + y − 1)(1 − x)(1 − y). Use this formula to find the maximum area of a triangle with perimeter 2.

Hint: You will need to solve a maximization problem

over a closed, bounded set D. To determine D, use the triangle inequality.

Kind of lost with where to begin. Any help and/or suggestions would be great! Thanks

and to find Maximum value of $A^2$ , you can use AM-GM inequality.
July 4th 2009, 10:01 PM
simplependulum

Use Heron's formula :

$s = 2/2 = 1$ and the other side is
$2 - x - y$

Therefore ,

$A = \sqrt{(s)(s-x)(s-y)[s-(2-x-y)]}$

$A^2 = (1)(1-x)(1-y)(1-2+x+y) = (1-x)(1-y)(x+y-1)$

Consider

$\frac{ (1-x) + (1-y) + (x+y-1) }{3} \geq [(1-x)(1-y)(x+y-1)]^{\frac{1}{3}}$

$(\frac{1}{3})^3 \geq A^2$

$A \leq \sqrt{\frac{1}{27}} = \frac{ \sqrt{3} }{9}$

so the maximum value of A is $\frac{ \sqrt{3} }{9}$

The equality holds when

$(1-x) = (1-y) = (x+y-1)$

it gives $x=y= \frac{2}{3}$ and the other side is also $\frac{2}{3}$ . It is an equilateral $\Delta$ .