# A pieces of wire 10cm long is cut in to two pieces, one pc. is bent into a square and the other is bent into an equalateral triangle.How shoul the wire be cut so that the total enclose area is:a)...

A pieces of wire 10cm long is cut in to two pieces, one pc. is bent into a square and the other is bent into an equalateral triangle.

How shoul the wire be cut so that the total enclose area is:

a) Maximum

b) minimum

NOTE: a+b+c=10-x

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You need to cut the wire into two pieces whose lengths are not given. You need to consider these lengths as `3x` and `4y` .

The problem provides the information that the total length of wire is of 10 cm such that:

`3x+4y = 10`

Consider that you need to use the length of `3x` to form the equilateral triangle and the length of `4y` to form the square.

You need to evaluate the area of equilateral triangle using the formula:

`A_(Delta) = l^2sqrt3/4`

Since the length of side of triangle is of `3x/3 = x` yields:

`A_(Delta) =x^2sqrt3/4`

You need to evaluate the area of the square using the formula:

`A_(square) = l^2`

Since the length of side of square is of `4y/4 =y` yields:

`A_(square) = y^2`

The area that needs to be optimized is:

`A = A_(Delta) + A_(square)`

`A = x^2sqrt3/4 + y^2`

You should write this equation in term of one variable, hence, you need to use the relation `x+y=10` to write y in terms of x such that:

`y = 10-x`

Substituting `10-x` for y in equation of area yields:

`A = x^2sqrt3/4 + (10-x)^2`

`A = x^2sqrt3/4 + 100 - 20x + x^2`

You should remember that the value of area is maximum or minimum at `A'(x) = 0` , hence, you need to differentiate the equation `A = x^2sqrt3/4 + 100 - 20x + x^2` with respect to x such that:

`A'(x) = xsqrt3/2 - 20 + 2x`

You need to solve the equation `A'(x) = 0` such that:

`xsqrt3/2 - 20 + 2x = 0 => 4x + xsqrt3 = 40`

Factoring out x yields:

`x(4 + sqrt3) = 40 => x = 40/(4 + sqrt3)`

Hence, you may find y substituting `40/(4 + sqrt3)` for x in equation `y = 10- x` such that:

`y = 10 - 40/(4 + sqrt3) => y = (40 + 10sqrt3 - 40)/(4 + sqrt3)`

`y = 10sqrt3/(4 + sqrt3)`

**Since, evaluating the second derivative, yields positive, then the found values `x = 40/(4 + sqrt3)` and `y = 10sqrt3/(4 + sqrt3)` minimize the area.**

**The problem with the above answer ***(by sciecesolve)* is that with the above values of x and y we need a wire length other than 10cm.

because **length of wire = 3x+4y**

Substituting the calculated values of x and y i.e. 40/(4+sqrt(3)) and 10*sqrt(3)/(4+sqrt(3)) we get:

**Length of wire** = 3*40/(4+sqrt(3))+4*10*sqrt(3)/(4+sqrt(3))

=>** (120+40*sqrt(3))/(4+sqrt(3)) cm > 10cm**

Thus the length of wire required according to the calculated values of x and y axceed the given length of wire 10cm.

In fact the areas of regular shapes increase for the same length of wire in sequence of an equilateral **triangle**, **square**, regular **pentagon**, regular **hexagon**, regular **heptagon**, regular **octagon** and so on till we get a **circle**.

The area of equilateral triangle is smallest and that of a circle is largest for the same perimeter.

Maximum or minimum area of a combination of two shapes cannot be achieved unless one of the shapes has a zero area with the same perimeter length.

The area is maximum when the wire is cut in length 10cm for square and Zero cm for triangle.

**The maximum area is** (10/4)^2+0 = 25/4cm^2 or **6.25cm^2**

The area is minimum when the wire is cut in length 0cm for square and 10cm for triangle.

**The minimum area is** 0+sqrt(3)*(10/3)^2/4 = **sqrt(3)*25/9cm^2**

Note: area of an equilateral triangle

I can not give any reference off the cuff but it is also true in case of volume which increases for a fixed surface area in sequence of tetrahedron, cube and sphere. I shall let you know the reference if I found one.

Thank you for you help. If you will mind, can I ask some reference in solving this kind of problem?? thanks..