The perimeter of an isosceles triangle is 18 cm. What should be its sides for the area of the triangle to be the largest?

1. Let’s designate:

2a – base of an isosceles triangle;
b – lateral side;
h is the height drawn to the base.
2. Let’s compose the equation for the perimeter of the triangle:

2a + 2b = 18;
a + b = 9;
b = 9 – a.
3. Find the height of the triangle using the Pythagorean theorem:

a ^ 2 + h ^ 2 = b ^ 2;
h ^ 2 = b ^ 2 – a ^ 2 = (9 – a) ^ 2 – a ^ 2 = 81 – 18a + a ^ 2 – a ^ 2 = 81 – 18a = 9 (9 – 2a);
h = √9 (9 – 2a) = 3√ (9 – 2a).
4. The area of ​​a triangle is equal to half the product of the base and the height:

S = 1/2 * 2a * h = ah = 3a√ (9 – 2a).

5. Find the maximum of the function f (a) = S ^ 2:

f (a) = 9a ^ 2 (9 – 2a) = 9 (9a ^ 2 – 2a ^ 3);
f ‘(a) = 9 (18a – 6a ^ 2) = 54a (3 – a) = 0.
a = 3 (cm) is the maximum point at which f (a), and hence S, takes the greatest value.

b = 9 – a = 9 – 3 = 6 (cm).

6. Sides of a triangle:

2a = 6 cm;
b = 6 cm.
Answer. The triangle must be equilateral.