An isosceles triangle has perimeter 64m. Find the dimensions of this triangle that make its area a maximum

To find the dimensions of an isosceles triangle that maximize its area given a perimeter of 64m, we can use the concept of optimization. Here's how you can solve the problem step by step:

Step 1: Understand the problem:
The problem states that we have an isosceles triangle, which means it has two sides of equal length. The perimeter of the triangle is given as 64m, and we need to find the dimensions (i.e., lengths of sides) that will maximize its area.

Step 2: Formulate the equation to optimize:
We know that the perimeter of an isosceles triangle is given by the formula:

Perimeter = 2a + b,
where 'a' is the length of the two equal sides, and 'b' is the base of the triangle.

In this case, the perimeter is given as 64m. So we have:

64 = 2a + b.

Step 3: Express the area in terms of a single variable:
The formula to calculate the area of a triangle is (base * height) / 2. In the case of an isosceles triangle, the height bisects the base, forming two right-angled triangles. Since we want to express everything in terms of a single variable, we can use the Pythagorean theorem to find the height.

Let x be the height of the triangle, and since it bisects the base, we have:

b = 2x.

The area (A) of the triangle can be expressed as:

A = (b * x) / 2 = (2x * x) / 2 = x^2.

So, we have now expressed the area (A) in terms of a single variable, x.

Step 4: Relate the variables:
We have two equations now:
64 = 2a + b, and
A = x^2.

We can rewrite the first equation in terms of 'b' using the relation b = 2x:
64 = 2a + 2x,
Divide both sides by 2:
32 = a + x,
Rearranging the equation gives:
a = 32 - x.

Now, we can rewrite the area equation using the value of 'a' we just derived:
A = x^2 = (32 - x)^2.

Step 5: Finding the maximum area:
To find the maximum value, we need to find the value of 'x' that maximizes the area. We can do this by taking the derivative of A with respect to x, setting it equal to zero, and solving for x.

Differentiating A = (32 - x)^2 with respect to x gives:
dA/dx = -2(32 - x) = 0,
Simplifying the expression:
64 - 2x = 0,
2x = 64,
x = 64/2,
x = 32.

So, we find x = 32m.

Step 6: Calculate the remaining dimensions:
Now that we have the value of x, we can substitute it back into our equations to find the remaining dimensions.
Using the first equation:
64 = 2a + b,
64 = 2a + 2x,
64 = 2a + 2(32),
64 = 2a + 64,
2a = 0,
a = 0.

Since 'a' represents the lengths of the two equal sides, if it equals 0, the triangle would be degenerate and not exist.

Therefore, there is no isosceles triangle with a maximum area that satisfies the given perimeter of 64m.