Maximizing the Area of an Isosceles Triangle with a Fixed Perimeter

In this article, we will explore a mathematical problem that involves finding the maximum possible area of an isosceles triangle with a fixed perimeter of 60 cm. This problem combines concepts from geometry and calculus to provide an in-depth understanding of maximizing areas under certain constraints. We will use a step-by-step approach to solve this problem, starting from the definition of the triangle's sides to the application of Heron's formula and the use of calculus to find the critical points.

Step 1: Define the Triangle's Sides

Let's denote the lengths of the two equal sides of the isosceles triangle as a and the base as b. The perimeter P of the triangle is given by:

P 2a b 60 cm

From this relationship, we can express b in terms of a:

b 60 - 2a

Step 2: Use Heron's Formula for Area

The area A of a triangle can be calculated using Heron's formula, which requires the semi-perimeter s of the triangle:

s frac{P}{2} 30 cm

Substituting the values, we get:

A sqrt{s(s - a)(s - a)(s - b)} sqrt{30(30 - a)(30 - a)(30 - (60 - 2a))}

Further simplifying, we get:

A sqrt{30(30 - a)(30 - a)(2a - 30)}

Step 3: Express the Area in Terms of Height

To simplify the process of finding the maximum area, let's use the alternative formula for the area of a triangle:

A frac{1}{2} times b times h

where h is the height from the apex to the base. Using the Pythagorean theorem to express h in terms of a and b:

h sqrt{a^2 - left(frac{b}{2}right)^2} sqrt{a^2 - left(frac{60 - 2a}{2}right)^2} sqrt{a^2 - (30 - a)^2}

Further simplifying:

h sqrt{a^2 - (900 - 60a a^2)} sqrt{60a - 900}

Step 4: Maximize the Area

The area can now be expressed as:

A frac{1}{2} times (60 - 2a) times sqrt{60a - 900}

To find the maximum area, we can recognize that when the triangle is equilateral, it achieves its maximum area for a given perimeter. Therefore, we set 2a b and solve for a:

2a a 60

3a 60

a 15 cm, hence b 30 cm

Step 5: Calculate the Maximum Area

For an equilateral triangle, the height h can be calculated as:

h frac{sqrt{3}}{2} times 30 15sqrt{3}

The area of the equilateral triangle is then:

A frac{1}{2} times 30 times 15sqrt{3} 225sqrt{3} text{cm}^2

Conclusion

The maximum possible area of an isosceles triangle with a perimeter of 60 cm is:

225sqrt{3} text{cm}^2 approx 389.74 text{cm}^2