Question

If two equal sides of an isosceles triangle have length a, find the length of the third side that maximizes the area of the triangle

Answer 1

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
`AC = a`. Let's call the height
`CD = x`. We can deduce the other leg:

`AD = √(AC^2-CD^2) = √(a^2-x^2)`

The area of ACD is given by half the product between the base and the height:

`A = \cfrac{AD \cdot CD}{2} = \cfrac{x√(a^2-x^2)}{2}`

and we want this quantity to be maximized. Again, it is sufficient to maximize the numerator.

To maximize it, let's compute its derivative:

`\left(x√(a^2-x^2)\right)' = \cfrac{(a^2 - 2 x^2)}{√(a^2 - x^2)}`

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

`a^2 - 2 x^2 = 0 \iff 2x^2 = a^2 \iff x^2 = \cfrac{a^2}{2} \iff x = \pm \cfrac{a}{√(2)}`

We can't accept a negative solution, so the answer will be

`x = \cfrac{a}{√(2)}`

Now, remember that the third side is twice AD, which means

`AB = 2AD = 2√(a^2-x^2) = \sqrt{a^2 - \cfrac{a^2}{2}} = \sqrt{\cfrac{a^2}{2}} = \cfrac{a}{√(2)}`

Answer 2

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