A piece of wire 10m long is cut into two pieces.?

so u have 2 lengths (which become the perimetre)

Length x and length (10-x)

x = perimetre of a square, therefore each side becomes length x/4, and the area of the square is always (x/4)^2

The triangle can be thought of as 2 right angled triangles as it is an equilateral, with the hypoteneuse being length (10-x)/3. the base (of one right angled triangle will be ((10-x)/3)/2

You can use pythagoras' to calculate the height of the triangle. (Rearrange for x)

((10-x)/3)^2 = (((10-x)/3)/2)^2 + height^2

total area therefore = (height x (10-x)/3) x 1/2) + (x/4)^2

Area of square is x²/16

A side of the triangle has length (10-x)/3 and its area is

√3(10-x)²/36 and total area is

A= x²/16+√3(10-x)²/36

This is a positive definite quadratic function of x ( i.e. is a U-shaped parabola

lying above the x-axis) and so has a minimum point where

dA/dx=0 giving

x/8 - √3(10-x)/18=0 and hence

x= 40√3/(9+4√3) which simplifies to

x=40(3√3-4)/11

Let the length of the wire vbe AC

AB=x so BC=10-x

Making a square of length x cm it means that each side is x/4

The other piece of the wire has a length of (10-x) cm

If this length is used to construct an equilateral triangle it means that each side will be (10-x)/3

Area of any triangle is (1/2) base times height

here the height h when we draw it , according to Pythagoras will be :

h^2+((10-x)^2)/9-((10-x)^2)/36 (The base (10-x)/2 divided by 2 is (10-x)/6!

so Area of triangle is : (1/2)((10-x)/3)((10-x)/sqrt(12))=(10-x)^2/(12sqrt(3))

Total area = area of square + area of triangle

A(T)=x^2/16+((10-x)^2)/(12sqrt(3))

differentiate A(T) for a minimum and equate to zero

dA/dx=x/8+((10-x)(-1))/6sqrt(3)=0

x=4/(4+sqrt(3)) mult. by 4-sqrt(3) to obtain

(16-12sqrt(3))/(16-27) so x=0.4 cm

A=(10-x)^2 + x^2 * sqrt(3) / 4

The minimum is there:

x = 6.9783052075