# 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) Maximumb) minimum                               NOTE: a+b+c=10-x

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...

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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.

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