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. How should the wire be cut

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. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) a minimum?

 

 

Ok, I get everything except how you determine a formula for the area of an equilateral triangle

 

According to the book:

 

A = x²/4 + (10 - x)²/12√3

 

I understand where the x²/4 and (10 - x) come from, but what about the 12/√3? And why is (10 - x) squared? After that, I understand how to find the minimum and maximum areas, but that little part tripped me up.

 

#30, practice test 3

Answer 1

If the wire is divided into x and 10 - x and the part assigned to x is bent into the square, then each side of the equilateral triangle will have a length of (10-x)/3.

 

The formula for the area of an equilateral triangle when you know a side "s" is A = s²√3/4 (this formula is easily proven by dropping the altitude from one vertex and using Pythagorean theorem).

 

So for this problem the area of the triangle is: A = [(10-x)/3]² (√3/4).

 

To see that this is the same as your book:

(10- x)²/9 (√3/4) = 

 

(10-x)²√3      √3                 (10 - x)² (3)             (10 -x)²

------------- -----------  =  -------------------- =      ------------

     36              √3                  36√3                      12√3

 

So, indeed a formula for the total area is:

 

A = (x/4)² + (10 - x)²/12√3

 

To find A' it's probably easiest to use the chain rule on each term.

 

Here is a start:

 

A' = 2(x/4)(1/4)  + 2(10-x)(-1)/12√3

 

Simplify a bit and set equal to zero, then solve for x. My guess is your book has an exact answer, i.e. an answer involving radical expression.

Answer 2

Suppose x is the cutting distance from one end of the wire.

Then, area of square would be ,

 

   A_sq =  (x/4)²

 

The side of equiladeral triangle would be,

 

  L=(10-x)/3 , its height would be (10-x)/3*sqrt(3)/2=(10-x)sqrt(3)/6  , so that, area of triangle is

 

A_tr = ((10-x)/3)²_* sqrt(3)/4=(10-x)² *sqrt(3)/36

 

So, the total area is ,   A = (x/4)² + (10 - x)²/36√3

 

This is a bit different than your book says ...

 

I bet either your book is wrong, or you're missing some information here. Otherwise,I am offering you the expression of Area as function of x - you find the dA/dx=0 and so on ...

 

message me if questions,

Good luck!    

 

dA/dx=x/8-2*sqrt(3)/36(10-x) = x/8+20*sqrt(3)/36-2*sqrt(3)/36*x=0

 

x(1/8+sqrt(3)/18)=10*sqrt(3)/18

 

x*0.22122 = 0.96225    x=4.35(m)

 

This corresponds to minimum area which is:

  A(4.35)=(4.35/4)^2+(10-4.35)^2*sqrt(3)/36=1.183+1.536=2.719 (m^2)

 

The maximum area is found cutting the wire at the very end - there derivative doesn't exist - it.s just extreme case with zero length and area of triangle and entire wire is spent on square - it's are would be (10/4)^2=6.25 (m^2)