.
A triangle has two sides of length 5. What length should be chosen for the third side
of the triangle so as to maximize the area within the triangle?
~~~~~~~~~~~~~~~~~~~~~
I will give two geometric solutions here.
It is clear that your triangle is isosceles.
Take an other copy (instance) of this triangle and attach these two triangles "base-to-base".
You will get a rhombus with the side length of 5 units.
Now ask yourself - which rhombus with the side length of 5 units has maximum area ?
Place such rhombus on its side to a horizontal line .
The rhombus is a deformable quadrilateral, so deform it to get maximum area.
It is clear that the maximum area is achieved when the rhombus is a square.
So, the maximum possible area of a rhombus is 5 x 5 = 25 square units,
and the diagonal of the rhombus is units then,
giving the optimum size to the third side of the triangle of units. ANSWER
Another possible solution is even simpler. See below.
The area of this triangle is half the product of the two given sides by the sine of the concluded angle
area of the triangle = .
is maximal when is the right angle. Then the area of the triangle is maximal
area of the triangle = = 12.5 square units,
and the opposite side "c" is, obviously, c = = .
Solved in two different ways, for your better understanding.