Consider the attached image. Let's focus on the triangle ACD. This is a right triangle, and is half of ABC. So, if we maximize the area of ACD, we also maximize the area of ABC.
We know that [tex] AC = a [/tex]. Let's call the height [tex] CD = x [/tex]. We can deduce the other leg:
[tex] AD = \sqrt{AC^2-CD^2} = \sqrt{a^2-x^2} [/tex]
The area of ACD is given by half the product between the base and the height:
[tex] A = \cfrac{AD \cdot CD}{2} = \cfrac{x\sqrt{a^2-x^2}}{2} [/tex]
and we want this quantity to be maximized. Again, it is sufficient to maximize the numerator.
To maximize it, let's compute its derivative:
[tex] \left(x\sqrt{a^2-x^2}\right)' = \cfrac{(a^2 - 2 x^2)}{\sqrt{a^2 - x^2}} [/tex]
We have a critical point where the derivative equals zero, and a fraction equals zero if and only if its numerator equals zero. So, the critical points are given by
[tex] a^2 - 2 x^2 = 0 \iff 2x^2 = a^2 \iff x^2 = \cfrac{a^2}{2} \iff x = \pm \cfrac{a}{\sqrt{2}} [/tex]
We can't accept a negative solution, so the answer will be
[tex] x = \cfrac{a}{\sqrt{2}} [/tex]
Now, remember that the third side is twice AD, which means
[tex] AB = 2AD = 2\sqrt{a^2-x^2} = \sqrt{a^2 - \cfrac{a^2}{2}} = \sqrt{\cfrac{a^2}{2}} = \cfrac{a}{\sqrt{2}} [/tex]
x = (2/√5)*a is the length of the third side which maximize the area
Isosceles triangle: ( See the Attached picture)
Triangle with two sides equal
In general the area of a triangle is A(t) = (1/2)* b*h where "b" is the base and "h" the height.
In picture attached
Area of triangle ABC = (1/2)* x*h (1)
And according to Pythagora´s theorem
a² = h² + (x/2)²
h² = a² - (x/2)²
h = √ a² - (x/2)²
By substitution in equation (1) we get A of the triangle as a function of x therefore:
A(x) = (1/2)*x*√ a² - (x/2)²
Tacking derivatives on both sides of the equation
A´ = (1/2) * √ a² - (x/2)² + ( - x) * (1/2)*x*] / √ a² - (x/2)²
A´ = (1/2) * √ a² - (x/2)² - (1/2) * x² / √ a² - (x/2)²
A´= 0
A´ = [(1/2) * √ a² - (x/2)² ]* √ a² - (x/2)² - (1/2) * x² = 0
A´ = a² - (x/2)² - x² = 0
a² - x²/4 - x² = 0
4a² - x² - 4*x² = 0
4*a² - 5*x² = 0
4*a² = 5*x²
2*a = √5 * x
x = (2/√5)*a
