For the calculus lovers! An isosceles triangle has perimeter of 64 cm. determine the side lengths if the triangle area is to be a maximum! HELPIts...

The perimeter of any geometric shape may be found adding the lengths of the sides of the geometric shape.

Since the isosceles triangle has two equal sides the perimeter is evaluated such that:

P = 2x + y

The perimeter is given such that 64 = 2x + y.

The area of the isosceles triangle is: `A = y*height/2`

The height may be found using the Pythagora's theorem:

`x^2 = y^2/2 ` + height

height = `x^2 - y^2/2`

Use the relation involving perimeter to write y in terms of x.

y = 64 - 2x

height =`x^2 - (64 - 2x)^2/2`

height = `x^2 - 4(32 - x)^2/2 =gt`  height =`x^2 - 4(32 - x)^2`

Area `= ((64 - 2x)*(x^2 - 4(32 - x)^2))/2`

Area `= (2*(32 - x)*(x^2 - 4(32 - x)^2))/2`

Area `= (32 - x)*(x^2 - 4(32 - x)^2))`

Expanding the binomial yields:

`A(x) =(32 - x)*(x^2 - 4*32^2+64*4*x - 4x^2)`

`A(x) =(32 - x)*(-3x^2 - 4*32^2+64*4*x)`

`A(x) =(32 - x)*(x^2 -4096+256x - 4x^2)`

`A(x) =32x^2 -131072+8192x - 128x^2 - x^3 + 4096x - 256x^2 + 4x^3`

`` `A(x) = 3x^3 - 352x^2 + 12288x - 131072`

The area is given by the equation `A(x) = 3x^3 - 352x^2 + 12288x - 131072.`

The Area is maximum if A'(x) = 0.

You need to differentiate the function of area:

`A'(x) = 9x^2 - 704x + 12288`

If `A'(x) = 0 =gt 9x^2 - 704x + 12288 = 0`

Applying quadratic formula yields: `x_(1,2) = (704+-sqrt(495616 - 442368))/18`

`` `x_(1,2) ~~ (704+-230)/18`

`x_1~~ 51.88 ; x_2~~ 26.33`

The area is maximum if the lengths of the sides of the isosceles triangle are: `x_1~~ 51.88 ; x_2~~ 26.33.`