can you guys pls help me out !

Question

A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

(a) Draw a diagram to illustrate the problem, using x as the variable to represent the length of one of the two pieces. Mark the sides of the triangle and the square in terms of x.

(b) Show that the total area of the square and the triangle can be represented as:

TA(x) = 9x² + 4√3x² −180x + 900/ 144

 

(c) How should the wire be cut so that the total area is a minimum?

Answer 1

--------      ------------

   x              10-x

 

If we take the piece of length x and

fold into equilateral triangle each 

side is length x/3

        / \

       /   \  x/3

      /      \

      -------

        x/3

 

area = (1/2)bh = (1/2)(x/3)height

 

to find height draw a vertical line up from

the center of the base to the top point to

create two right triangles... 

legs are x/6 and h, hypotenuse = x/3

        |\

     h |  \ x/3

        |__\.       right half of the triangle

         x/6

From pythagorean theorem 

(x/6)²+h² (x/3)²

h²= (x/3)²-(x/6)² = x²/9 - x²/36

h² = 3x²/36

h = √(3x²/36)

h = x√3/6

 

area of triangle = (1/2)(x/3)(x√3/6)

= (√3)x²/36

 

For the square:  If we bend a length of

10-x into a square we have four sides of

length (10-x)/4

 

Area of the square:  [(10-x)/4]² = (x²-20x+100)/16

 

T(A) = triangle area plus square area

T(A) = (√3 x²/36)+(x²-20x+100)/16

 

lowest common denominator for 36 and 16 is 144

 

T(A) = (4√3 x²)/144 + 9(x²-20x+100)/144

T(A)= [(4√3)x² +9x²-180x+900]/144

 

For minimum area take the quadratic equation

we have created. a=4√3+9, b=-180, c=900

 

The parabola opens downward so minimum area is

at the vertex where x = -b/2a

x = 180/2(4√3+9) = 5.65m

 

Cut the triangle piece to 5.65m

Cut the square piece to 4.35m