Optimization Math Help: Finding Minimum Enclosed Area with Wire Cut

[fghtffyrdmns]

Homework Statement

A piece of wire 8 cm long is cut into two pieces. One piece is bent to form a circle, and the other is bent to form a square. How should the wire be cut if the total enclosed area is to be small as possible? Keep \pi in your answer.

Homework Equations

A= \pi r^{2}
A= lw

The Attempt at a Solution

this is the part I am confused about. I cannot seem to set up the question. I do now that I have to find the function, take the derivative and equal it to zero (use 2nd derivative test to check that it's a min), and solve.

 

[Donaldos]

Let l be the length of the piece that is used to form a square.

Prove that :

A(l)=\frac{l^2} {16} +\frac{\left(8-l\right)^2}{4\pi}

Then you can derivate that expression.

 

[fghtffyrdmns]

Let l be the length of the piece that is used to form a square.

Prove that :

A(l)=\frac{l^2} {16} +\frac{\left(8-l\right)^2}{4\pi}

Then you can derivate that expression.

Whoah, where did that come from?

I was going to say, can't you just treat the square and circle separately?

 

[Mark44]

Whoah, where did that come from?

I was going to say, can't you just treat the square and circle separately?

No, since you have to find the location to cut the wire so that you get the maximum area for both figures.

 

[fghtffyrdmns]

What if you make x be the length of the wire?

Then you get A= (8-x)^{2} for the area of the square

 

[Mark44]

That won't work, since you're assuming that all four sides are 8 - x cm long.

If your wire extends from 0 to 8 cm, and you cut it at x, that gives you two pieces of wire. One of them is x cm long and the other is 8 - x cm long. One piece will be used to form a square - all four sides. The other piece will be used to form a circle.

So which piece will you use to make the square, and which piece will you use to make the circle? Keep in mind that each piece will form the perimeter of whatever object you make out of it.

 

[fghtffyrdmns]

I still cannot understand this question : (

Let x be the lentgh

thus should you not get 8-x?

then you can plug it into the area of a cirlce equation to get pi(8-x)^{2}?

 

[Mark44]

then you can plug it into the area of a cirlce equation to get pi(8-x)^{2}?

No, not at all.

Try to be more specific. x defines the point at which the 8 cm wire is cut.

You will end up with one piece that is x cm long, and another that is 8 - x cm long. You're going to bend these pieces to form a square and a circle. Is it reasonable that the circle's radius is 8 - x cm? That's the formula that you seem eager to use.

 

[fghtffyrdmns]

No, not at all.

Try to be more specific. x defines the point at which the 8 cm wire is cut.

You will end up with one piece that is x cm long, and another that is 8 - x cm long. You're going to bend these pieces to form a square and a circle. Is it reasonable that the circle's radius is 8 - x cm? That's the formula that you seem eager to use.

So 8 - x is the square while x is the square?

 

[Mark44]

So 8 - x is the square while x is the square?

Huh?

 

[fghtffyrdmns]

Huh?

I mean x-8 is the square while x is the circle.

I can't think :(

 

[fghtffyrdmns]

I got A(x) = x^{2}+\pi(8-x)^{2}

 

[Mark44]

I got A(x) = x^{2}+\pi(8-x)^{2}

Not even close.

You have the two pieces of wire: x cm for the square and 8 - x cm for the circle. Describe to me in words exactly how you will form the two figures from these pieces of wire.

 

[Office_Shredder]

It might help if you actually take a piece of string or something and try to make a circle out of it. What does the length of the string represent here?

 

[fghtffyrdmns]

It might help if you actually take a piece of string or something and try to make a circle out of it. What does the length of the string represent here?

Circumference.

 

[Mark44]

Right - not the radius. Good.

And for the square, what does the length of that string represent?

So if you bend the x cm piece to make a square and the 8 - x piece to make a circle (BTW, it might be easier to use the x piece for the circle and the 8 - x piece for the square, what are the dimensions of each figure?

 

[fghtffyrdmns]

Right - not the radius. Good.

And for the square, what does the length of that string represent?

So if you bend the x cm piece to make a square and the 8 - x piece to make a circle (BTW, it might be easier to use the x piece for the circle and the 8 - x piece for the square, what are the dimensions of each figure?

the length is is each side. If I make (8-x) for the square, I would just get 8x-x^{2}

I don't seem to remember how to get area with circumference.

 

[Mark44]

the length is is each side. If I make (8-x) for the square, I would just get 8x-x^{2}

How did you get that? What are the dimensions of the square?

I don't seem to remember how to get area with circumference.

I'm not asking for area. I'm asking for the dimensions of the circle, namely its radius.

 

[Mark44]

I think it would be useful for you to take a wire coathanger and cut a piece out of it 8 inches long. Then cut your 8" long piece into two pieces, say 3" and 5". Take one of them and form it into a square and form the other into a circle. If you use the 3" piece to make a square, what are the lengths of the each side of the square? If you use the 5" piece for a circle, what is the radius of the circle.

Once you know the dimensions of each object, then you can calculate the area of each, and hence, the total area.

 

[fghtffyrdmns]

You can't tell the dimensions though. It doesn't say it was cut in half.

 

[Office_Shredder]

You take a piece of wire that is of length 8-x inches. You bend it into a square. What is the length of one side of the square? This question is definitely answerable. Do two steps:
1) What is the perimeter of this square?
2) Based on 1), what is the length of a side of this square?

Once you have done that you can calculate the area

 

[fghtffyrdmns]

You take a piece of wire that is of length 8-x inches. You bend it into a square. What is the length of one side of the square? This question is definitely answerable. Do two steps:
1) What is the perimeter of this square?
2) Based on 1), what is the length of a side of this square?

Once you have done that you can calculate the area

The length of the square is 8-x? Then you get 32 - 4x for perimeter

I am not following this.

 

[Office_Shredder]

If I have a piece of wire that is 8-x inches long, and I bend it into a square, you DO NOT get that each side is 8-x inches long. The perimeter will be 8-x inches long

 

[fghtffyrdmns]

If I have a piece of wire that is 8-x inches long, and I bend it into a square, you DO NOT get that each side is 8-x inches long. The perimeter will be 8-x inches long

Oh, I thought you said that's the length. I was wondering what's up. I know that the area of the square is going to be \frac{(8-x)^{2}}{16}. It's the circle that I cannot seem to get.

Edit: ahhh, I got it now. \frac{x^{2}}{4\pi}

I solved for the diameter using the circumference formula then divided by 2 get the radius.

 

[fghtffyrdmns]

Wait, if the perimeter is 8-x, wouldn't each side be 8-x/4?


<br /> \frac{x^{2}}{4\pi}<br />

is the area of the circle.

 

[Mark44]

OK, good. to summarize, the length of each side of the square is (8 - x)/4, and the radius of the circle is x/(2*pi). The combined area of the two figures is A(x) = (8 - x)²/16 + x²/(4 * pi).

Now if I can remember back to the original problem, you wanted to maximize the total area. Can you do that?

 

[Mark44]

Wait, if the perimeter is 8-x, wouldn't each side be 8-x/4?


<br /> \frac{x^{2}}{4\pi}<br />

is the area of the circle.

Each side would be (8 - x)/4 -- you need parentheses when you write it on one line, and yes, that's the area.

 

[Office_Shredder]

Each side would be 1/4th the perimeter so (8-x)/4 inches. So the area would be (8-x)²/16.

So what is your formula for the total area based on x?

 

[fghtffyrdmns]

OK, good. to summarize, the length of each side of the square is (8 - x)/4, and the radius of the circle is x/(2*pi). The combined area of the two figures is A(x) = (8 - x)²/16 + x²/(4 * pi).

Now if I can remember back to the original problem, you wanted to maximize the total area. Can you do that?

Yes, sir :).

However, I have a math-legality question about that. Since I'm diving by 16 and 4pi, can I factor it out then just take the derivative of (8-x)^2 and x^2? Or must I use the quotient rule on both?

 

[Mark44]

You could use the quotient rule if you don't mind working harder than you need to. For example, you could use the quotient rule to differentiate f(x) = x/4, but that would be silly. Instead, you could notice that x/4 = (1/4)x, so d/dx(x/4) = d/dx(1/4*x) = 1/4 * d/dx(x), using the constant-multiple rule.

In your case d/dx[(8 - x)²/16 + x²/(4pi)] = (1/16) d/dx(8 - x)²) + 1/(4pi)*d/dx(x²). (I also used the sum rule for derivatives.)

 

[fghtffyrdmns]

You could use the quotient rule if you don't mind working harder than you need to. For example, you could use the quotient rule to differentiate f(x) = x/4, but that would be silly. Instead, you could notice that x/4 = (1/4)x, so d/dx(x/4) = d/dx(1/4*x) = 1/4 * d/dx(x), using the constant-multiple rule.

In your case d/dx[(8 - x)²/16 + x²/(4pi)] = (1/16) d/dx(8 - x)²) + 1/(4pi)*d/dx(x²). (I also used the sum rule for derivatives.)

Yes, that is what I mean :)

I got x = --
Edit: wrong answer.

 

[Mark44]

We have A(x) = (1/16)(8 - x)² + (1/(4pi)x².
A'(x) = ?

Your answer should be an equation.

 

[fghtffyrdmns]

We have A(x) = (1/16)(8 - x)² + (1/(4pi)x².
A'(x) = ?

Your answer should be an equation.

A'(x) = \frac{-16+2x}{16} + \frac{x}{2\pi}

However, I have to set the first derivative to 0 so I can find at which x value it should be cut so the area is small as possible (critical point - max/min value).

 

[Mark44]

Your A'(x) is correct, but it would be better to simplify it. You can rewrite it as A'(x) = (x - 8)/8 + x/(2pi).

There is also a domain to consider, since A(x) is not defined for all real numbers. For example, even though you can calculate a value for A(-8), you can't cut the wire at -8 cm.

Your critical point will come at a point where A'(x) = 0 or at an endpoint of the domain. What is the implied domain for A(x) = (1/16)(8 - x)² + (1/(4pi)x²?

 

[fghtffyrdmns]

0<x<8 (cannot be equal to since it makes no sense to cut 8 cm or 0 cm.) X is the element of any real number is the domain.

So I can't solve now?

Edit: 0 and 8 do not work .

 

[Mark44]

You can choose to not make the cut at all, so that you make only a square or only a circle. So the domain of A(x) is {x in the real numbers| 0 <= x <= 8}.

Now finish the problem. Be sure to check x = 0 and x = 8 after you have solved A'(x) = 0. That way you can be sure you have found the absolute maximum area.