An isosceles triangle is to have a perimeter of 64cm. Determine the side lengths of the triangle if the area is to be maximum.

I need help setting this question up! Please

To set up the problem, we first need to establish some variables and equations. Let's denote the length of the equal sides of the isosceles triangle as "x" and the base as "y".

Since it's mentioned that the perimeter of the triangle is 64cm, we can set up the following equation:
2x + y = 64 (equation 1)

To maximize the area of the triangle, we need to find the value of "x" and "y" that will give us the maximum possible area. The formula for the area of a triangle is A = 0.5 * base * height.

In an isosceles triangle, the height is the perpendicular line from the vertex opposite the base to the base itself. This line divides the triangle into two congruent right-angled triangles. To find the height, we can use the Pythagorean theorem.

The Pythagorean theorem states that in any right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In our case, the hypotenuse is x (the equal side length) and the other two sides are y/2 (half of the base) and the height (h). So, we can set up the following equation based on the Pythagorean theorem:
h^2 = x^2 - (y/2)^2 (equation 2)

Now, with equations 1 and 2, we have a system of two equations with two unknowns (x and y). To solve for x and y, we will use the method of substitution.

First, solve equation 1 for y:
y = 64 - 2x

Next, substitute this value of y into equation 2:
h^2 = x^2 - ((64 - 2x)/2)^2
Simplify the equation further:
h^2 = x^2 - (32 - x)^2

Now we have a single equation with one variable, h. To find the maximum area, we need to maximize h. To do this, we can find the critical points of the equation by taking the derivative of h^2 with respect to x and setting it equal to zero. Then, solve for x using calculus techniques.

Once we find the value of x, we can substitute it back into the equation 1 to find the corresponding value of y.