How should the wire be cut so that it is (a) min (b) max??

[TONYYEUNG]

Dear All,

Please go to the following to have a look of the 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) minimal (b) maximal??

Cheers,

Tony

 

[pka]

Where are you having problems?
What do you need to know about this problem?
What have you done on it?

 

[TchrWill]

Re: How should the wire be cut so that it is (a) min (b) max

Please go to the following to have a look of the 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) minimal (b) maximal??

\

You are dealing with a square and a triangle.

The minimuj area associated with both the square and triangle is as follows.

The area pf the square is x^2 and the area of the triangle is [.866(10 - 4x)(10 - 4x)/2]

Therefore, we can write x^2 +[.866(10 - 4x)(10 - 4x)/2] = 10

I'll let you expand this and take the first derivitive to determine the maximum area enclosable by the wire.

The maximum area eclosable would be to cut a miniscule piece off of the wire and make a square out f what is left.

The full length wire makes a squae with sides of 2.5 and an area of 6.26. The triangle has sides of 3.333...and an area of 4.81. Therefore, cutting off a small piece of the wire will produce a square with the maximum enclosed area.

 

[TONYYEUNG]

Dear pka,

I've tried to make the following equation but I am not sure it is right or not:-

Let x be the cutting point

A(x)=(x/4)(x/4)+((10-x)/4)((10-x)/4)

If the above equation is right. How can I do the next step? If not, could you tell me something about it??

 

[TONYYEUNG]

Dear TchrWill,

Thank you for your reply. I will try to solve the problem by some of your information.

Cheers,

Tony

 

[pka]

Tony, I like you approach best: A(x)=(x/4)(x/4)+((10-x)/4)((10-x)/4)
However, is s is the length of the side of an equilateral triangle the its area is: \(\displaystyle \L
\frac{{\sqrt 3 s^2 }}{4}.\)
Thus you want to use this expression: \(\displaystyle \L
A(x) = \frac{{\sqrt 3 }}{4}\left( {\frac{x}{3}} \right)^2 + \left( {\frac{{10 - x}}{4}} \right)^2\)