Wire cut to shape circle and square so total area is maximum or minimum

[5hassay]

Homework Statement

A piece of wire 40 cm long is cut into two pieces. One piece is bent into the shape of a square and the other is bent into the shape of a circle. How should the wire be cut so that the total area enclosed is a (a) maximum and (b) minimum?

Spoiler

The answer to part (a) is, "all wire for the circle," and for part (b), approximately 22.4 cm for the square.

Homework Equations

Area of a circle $= A_{c} = \pi r^{2}$.
Circumference of a circle $= C = 2 \pi r$.

Area of a square $= A_{s} = s^{2}$.
Perimeter of a square $= P = 4s$.

The Attempt at a Solution

I recognize that I will be needed to differentiate the area $A$ of both the square and circle combined. So, I then begin to construct an equation for the area enclosed by both geometric shapes.

Here, I began by reasoning that,

$40 = P + C = 4s + 2 \pi r$

and then solving for either $r$ or $s$ and substituting them into the equation,

$A = A_{s} + A_{c} = s^{2} + \pi r^{2}$

However, that did not seem to get me anywhere. (Why is that?)

Then, after some research, I changed my perspective. Specifically, I began reasoning that $l$ centimeters of length would be cut from the 40 cm wire and used to shape the square, and the remaining length of the wire, that is, $40 - l$, is used to shape the circle. This ended up correctly answering part (b), and the following is a summary of my work.

$P = l \Leftrightarrow 4s = l \rightarrow s = \frac{l}{4}$
$\Rightarrow A_{s} = (\frac{l}{4})^{2}$

$C = 40 - l \Leftrightarrow 2 \pi r = 40 - l \rightarrow r = \frac{40 - l}{2 \pi}$
$\Rightarrow A_{c} = \pi (\frac{40 - l}{2 \pi})^{2} = \frac{(40 - l)^{2}}{4 \pi}$

$\Rightarrow A = \frac{l^{2}}{16} + \frac{(40 - l)^{2}}{4 \pi}$

$\frac{d A}{d l} = \frac{1}{8} l - \frac{1}{2 \pi} (40 - l) \Leftrightarrow \frac{d A}{d l} = (\frac{1}{8} + \frac{1}{2 \pi}) l - \frac{20}{\pi}$

$\frac{d A}{d l} > 0 \Leftrightarrow l > \frac{\frac{20}{\pi}}{\frac{1}{8} + \frac{1}{2 \pi}} \Leftrightarrow l > \frac{\frac{20}{\pi}}{\frac{\pi + 4}{8 \pi}} \rightarrow l > \frac{160}{\pi + 4} \approx 22.4 cm$
$\frac{d A}{d l} < 0 \Leftrightarrow l < \frac{160}{\pi + 4}$

Therefore, there is a minimum at $l = \frac{160}{\pi + 4}$ -- for the square.

Now, for the maximum, I thought I would apply the same reasoning. That is, instead, the circle has a perimeter of $l$ and the square, $40 - l$. But, in performing this, I just got another minimum.

In observing the answer (see spoiler) for part (a), I imagine that there are actually four cases in which the wire could be used: starting with the square, then the circle, the reverse of that, using it all for the circle or all for the square. Of course, this adds confusion, as the question specifically states that the wire is used to shape a square and a circle, so there must be an error there if an answer was such that a square was not formed.

So, I went about using the entirety of the length of the wire for the perimeter of the circle and went about differentiating its area to see if a maximum occurs. Unfortunately, in the end, I did not get a maximum. Thus, I am maybe thinking that you were not expected to take a derivative, but simply calculate the area of the circle with a circumference of 40 cm and compare that with that of the square and see which was larger. But, I don't really like these methods, as I feel there should be one equation one can make which can be differentiated and set equal to zero thus it has multiple zero's; one a maximum, one a minimum. Then again, I just thought that other max./min. problems just involving a given equation $f(x)$ and you were to find the absolute maxima and or minima. Continuing, they often had endpoints you would use to see if the derivative's roots were absolute or local, or if the endpoints themselves were critical numbers. So, in relation to this problem, could the cases of using all 40 cm of wire length for one shape (despite the mentioned possible conflict) be thought of endpoints?

I would greatly appreciate any guidance on this problem.

 

[micromass]

Homework Statement

A piece of wire 40 cm long is cut into two pieces. One piece is bent into the shape of a square and the other is bent into the shape of a circle. How should the wire be cut so that the total area enclosed is a (a) maximum and (b) minimum?

Spoiler

The answer to part (a) is, "all wire for the circle," and for part (b), approximately 22.4 cm for the square.

Homework Equations

Area of a circle $= A_{c} = \pi r^{2}$.
Circumference of a circle $= C = 2 \pi r$.

Area of a square $= A_{s} = s^{2}$.
Perimeter of a square $= P = 4s$.

The Attempt at a Solution

I recognize that I will be needed to differentiate the area $A$ of both the square and circle combined. So, I then begin to construct an equation for the area enclosed by both geometric shapes.

Here, I began by reasoning that,

$40 = P + C = 4s + 2 \pi r$

and then solving for either $r$ or $s$ and substituting them into the equation,

$A = A_{s} + A_{c} = s^{2} + \pi r^{2}$

However, that did not seem to get me anywhere. (Why is that?)

You should have obtained the correct answer this way too. So I'm guessing you did something wrong...

Then, after some research, I changed my perspective. Specifically, I began reasoning that $l$ centimeters of length would be cut from the 40 cm wire and used to shape the square, and the remaining length of the wire, that is, $40 - l$, is used to shape the circle. This ended up correctly answering part (b), and the following is a summary of my work.

$P = l \Leftrightarrow 4s = l \rightarrow s = \frac{l}{4}$
$\Rightarrow A_{s} = (\frac{l}{4})^{2}$

$C = 40 - l \Leftrightarrow 2 \pi r = 40 - l \rightarrow r = \frac{40 - l}{2 \pi}$
$\Rightarrow A_{c} = \pi (\frac{40 - l}{2 \pi})^{2} = \frac{(40 - l)^{2}}{4 \pi}$

$\Rightarrow A = \frac{l^{2}}{16} + \frac{(40 - l)^{2}}{4 \pi}$

$\frac{d A}{d l} = \frac{1}{8} l - \frac{1}{2 \pi} (40 - l) \Leftrightarrow \frac{d A}{d l} = (\frac{1}{8} + \frac{1}{2 \pi}) l - \frac{20}{\pi}$

You want to find the minimum/maximum of A. But the candidates for the minimum and the maximum of A are not only the points where $\frac{dA}{dl}=0$. It are also the end points of the interval.

So your candidates for the minimum and the maximum are 0, 22.4 and 40. Now, which of these values corresponds to the minimum, maximum??

$\frac{d A}{d l} > 0 \Leftrightarrow l > \frac{\frac{20}{\pi}}{\frac{1}{8} + \frac{1}{2 \pi}} \Leftrightarrow l > \frac{\frac{20}{\pi}}{\frac{\pi + 4}{8 \pi}} \rightarrow l > \frac{160}{\pi + 4} \approx 22.4 cm$
$\frac{d A}{d l} < 0 \Leftrightarrow l < \frac{160}{\pi + 4}$

Therefore, there is a minimum at $l = \frac{160}{\pi + 4}$ -- for the square.

Now, for the maximum, I thought I would apply the same reasoning. That is, instead, the circle has a perimeter of $l$ and the square, $40 - l$. But, in performing this, I just got another minimum.

You just changed the parameters. I don't see why you should find a maximum instead of a minimum this time?

In observing the answer (see spoiler) for part (a), I imagine that there are actually four cases in which the wire could be used: starting with the square, then the circle, the reverse of that, using it all for the circle or all for the square. Of course, this adds confusion, as the question specifically states that the wire is used to shape a square and a circle, so there must be an error there if an answer was such that a square was not formed.

So, I went about using the entirety of the length of the wire for the perimeter of the circle and went about differentiating its area to see if a maximum occurs. Unfortunately, in the end, I did not get a maximum. Thus, I am maybe thinking that you were not expected to take a derivative, but simply calculate the area of the circle with a circumference of 40 cm and compare that with that of the square and see which was larger. But, I don't really like these methods, as I feel there should be one equation one can make which can be differentiated and set equal to zero thus it has multiple zero's; one a maximum, one a minimum. Then again, I just thought that other max./min. problems just involving a given equation $f(x)$ and you were to find the absolute maxima and or minima. Continuing, they often had endpoints you would use to see if the derivative's roots were absolute or local, or if the endpoints themselves were critical numbers. So, in relation to this problem, could the cases of using all 40 cm of wire length for one shape (despite the mentioned possible conflict) be thought of endpoints?

I would greatly appreciate any guidance on this problem.

 

[Ray Vickson]

Homework Statement

A piece of wire 40 cm long is cut into two pieces. One piece is bent into the shape of a square and the other is bent into the shape of a circle. How should the wire be cut so that the total area enclosed is a (a) maximum and (b) minimum?

Spoiler

The answer to part (a) is, "all wire for the circle," and for part (b), approximately 22.4 cm for the square.

Homework Equations

Area of a circle $= A_{c} = \pi r^{2}$.
Circumference of a circle $= C = 2 \pi r$.

Area of a square $= A_{s} = s^{2}$.
Perimeter of a square $= P = 4s$.

The Attempt at a Solution

I recognize that I will be needed to differentiate the area $A$ of both the square and circle combined. So, I then begin to construct an equation for the area enclosed by both geometric shapes.

Here, I began by reasoning that,

$40 = P + C = 4s + 2 \pi r$

and then solving for either $r$ or $s$ and substituting them into the equation,

$A = A_{s} + A_{c} = s^{2} + \pi r^{2}$

However, that did not seem to get me anywhere. (Why is that?)

Then, after some research, I changed my perspective. Specifically, I began reasoning that $l$ centimeters of length would be cut from the 40 cm wire and used to shape the square, and the remaining length of the wire, that is, $40 - l$, is used to shape the circle. This ended up correctly answering part (b), and the following is a summary of my work.

$P = l \Leftrightarrow 4s = l \rightarrow s = \frac{l}{4}$
$\Rightarrow A_{s} = (\frac{l}{4})^{2}$

$C = 40 - l \Leftrightarrow 2 \pi r = 40 - l \rightarrow r = \frac{40 - l}{2 \pi}$
$\Rightarrow A_{c} = \pi (\frac{40 - l}{2 \pi})^{2} = \frac{(40 - l)^{2}}{4 \pi}$

$\Rightarrow A = \frac{l^{2}}{16} + \frac{(40 - l)^{2}}{4 \pi}$

$\frac{d A}{d l} = \frac{1}{8} l - \frac{1}{2 \pi} (40 - l) \Leftrightarrow \frac{d A}{d l} = (\frac{1}{8} + \frac{1}{2 \pi}) l - \frac{20}{\pi}$

$\frac{d A}{d l} > 0 \Leftrightarrow l > \frac{\frac{20}{\pi}}{\frac{1}{8} + \frac{1}{2 \pi}} \Leftrightarrow l > \frac{\frac{20}{\pi}}{\frac{\pi + 4}{8 \pi}} \rightarrow l > \frac{160}{\pi + 4} \approx 22.4 cm$
$\frac{d A}{d l} < 0 \Leftrightarrow l < \frac{160}{\pi + 4}$

Therefore, there is a minimum at $l = \frac{160}{\pi + 4}$ -- for the square.

Now, for the maximum, I thought I would apply the same reasoning. That is, instead, the circle has a perimeter of $l$ and the square, $40 - l$. But, in performing this, I just got another minimum.

In observing the answer (see spoiler) for part (a), I imagine that there are actually four cases in which the wire could be used: starting with the square, then the circle, the reverse of that, using it all for the circle or all for the square. Of course, this adds confusion, as the question specifically states that the wire is used to shape a square and a circle, so there must be an error there if an answer was such that a square was not formed.

So, I went about using the entirety of the length of the wire for the perimeter of the circle and went about differentiating its area to see if a maximum occurs. Unfortunately, in the end, I did not get a maximum. Thus, I am maybe thinking that you were not expected to take a derivative, but simply calculate the area of the circle with a circumference of 40 cm and compare that with that of the square and see which was larger. But, I don't really like these methods, as I feel there should be one equation one can make which can be differentiated and set equal to zero thus it has multiple zero's; one a maximum, one a minimum. Then again, I just thought that other max./min. problems just involving a given equation $f(x)$ and you were to find the absolute maxima and or minima. Continuing, they often had endpoints you would use to see if the derivative's roots were absolute or local, or if the endpoints themselves were critical numbers. So, in relation to this problem, could the cases of using all 40 cm of wire length for one shape (despite the mentioned possible conflict) be thought of endpoints?

I would greatly appreciate any guidance on this problem.

You want to optimize A(x) = x^2/16 + (40-x)^2/(4*pi) over the interval 0 <= x <= 40; here, my x = your l. Setting the derivative to zero gives an interior optimum (a min in this case), but one also needs to look at the endpoints, where *the derivative need not vanish*. We have A(0) = 40^2/(4*pi) = 127.324, while A(40) = 40^2/16 = 100, so A(40) is the max. Note: for an endpoint we have: deriv >= 0 at left end for a local min, deriv <= 0 at left end for a local max; opposite signs are needed at the right-hand endpoint. So, x = 0 is the global max while x = 40 is a local max. Basically, in this problem there is no nice formula for the max; you need to examine several cases numerically.

RGV

 

[5hassay]

You should have obtained the correct answer this way too. So I'm guessing you did something wrong...



You want to find the minimum/maximum of A. But the candidates for the minimum and the maximum of A are not only the points where $\frac{dA}{dl}=0$. It are also the end points of the interval.

So your candidates for the minimum and the maximum are 0, 22.4 and 40. Now, which of these values corresponds to the minimum, maximum??



You just changed the parameters. I don't see why you should find a maximum instead of a minimum this time?

You want to optimize A(x) = x^2/16 + (40-x)^2/(4*pi) over the interval 0 <= x <= 40; here, my x = your l. Setting the derivative to zero gives an interior optimum (a min in this case), but one also needs to look at the endpoints, where *the derivative need not vanish*. We have A(0) = 40^2/(4*pi) = 127.324, while A(40) = 40^2/16 = 100, so A(40) is the max. Note: for an endpoint we have: deriv >= 0 at left end for a local min, deriv <= 0 at left end for a local max; opposite signs are needed at the right-hand endpoint. So, x = 0 is the global max while x = 40 is a local max. Basically, in this problem there is no nice formula for the max; you need to examine several cases numerically.

RGV

Thanks!

I do now understand, and everything is a lot more clear.

Once again, thank you both for your help.