Hello everyone,

One of my Calculus students had an interesting Related Rates problem that I had to go home and think about for a while in order to figure out. The problem was set up as such:

A 25 inch piece of rope needs to be cut into 2 pieces to form a square and a circle. How should the rope be cut so that the combined surface area of the circle and square is as small as possible?

One of my Calculus students had an interesting Related Rates problem that I had to go home and think about for a while in order to figure out. The problem was set up as such:

A 25 inch piece of rope needs to be cut into 2 pieces to form a square and a circle. How should the rope be cut so that the combined surface area of the circle and square is as small as possible?

Here's what we'll need to do:

1. We will have to form equations that relate the length of the perimeter and circumference to the combined surface area.

2. We will then differentiate to create an equation with the derivative of the surface area with respect to lengths of rope.

3. Wherever this derivative equals 0 there will be a maxima or minima, and so we will set the derivative = to 0 and determine which critical points are minima.

Let's start by looking at the problem conceptually and then defining some variables.

We are going to cut a piece of rope, and from the two pieces we will make a circle and a square. So we know then that:

Total length of rope = circumference of circle + perimeter of square

We are given the total length of rope in the problem and we know the formulas for circumference and perimeter:

25 in = 2πr + 4s

Where r is the radius of the circle and s is one of the sides of the square.

We're off to a good start. We have related the total length of rope to generic formulas for the circumference and perimeter, but of course, ultimately, we will want to only be left with two variables: one to represent the area and the other to represent the lengths of rope.

How can we represent the lengths of rope with just one variable?

Let's take a second look at the conceptual setup for this first equation:

Total length of rope = circumference of circle + perimeter of square

25 in = 2πr + 4s

Let's define the variable "x" to represent the length of the circumference of the circle. Whatever rope we use for the circle, the remainder of the 25" of rope is equal to the perimeter of the square. In other words, the perimeter of the square = 25 - x. Therefore:

circumference = x = 2πr

perimeter = 25 - x = 4s

perimeter = 25 - x = 4s

Now we have equations that relate the length of our rope to a variable which represents the circumference, and also to the sides of the square and the radius of the circle.

Remember that we are trying to determine the rate of change of the combined surface area to the length of the circumference, so we must now form another equation which will accomplish this for us. As we did before, let's think conceptually first:

Combined surface area = area of circle + area of square

Ok, seems pretty simple so far. Surface area is the value we're trying to find, so we'll define a variable "A" to represent surface area, and of course we know how to obtain the areas of a circle and a square:

A = πr

Aha! Now we have an equation which relates area to the radius of the circle and a side of the square, and previously we put together two equations which related radius and side to x, the variable we've set to represent the circumference of the circle.

Since we have now formed an equation relating r and s to A, if we use our previously constructed equations relating x to r and s, we can use substitution to directly relate A to x.

Recall from above that:

circumference = x = 2πr

perimeter = 25 - x = 4s

If 2πr = x, then:

Combined surface area = area of circle + area of square

Ok, seems pretty simple so far. Surface area is the value we're trying to find, so we'll define a variable "A" to represent surface area, and of course we know how to obtain the areas of a circle and a square:

A = πr

^{2}+ s^{2}Aha! Now we have an equation which relates area to the radius of the circle and a side of the square, and previously we put together two equations which related radius and side to x, the variable we've set to represent the circumference of the circle.

Since we have now formed an equation relating r and s to A, if we use our previously constructed equations relating x to r and s, we can use substitution to directly relate A to x.

Recall from above that:

circumference = x = 2πr

perimeter = 25 - x = 4s

If 2πr = x, then:

r = x / 2π

And if 25 - x = 4s, we can easily solve for s as follows:

And if 25 - x = 4s, we can easily solve for s as follows:

s = (25 - x) / 4

Perfect! Looking again at our equation for area:

A = πr

^{2}+ s

^{2}

All we need to do is substitute in for r and s to leave an equation relating A to x:

A = π( x / 2π )

^{2}+ ( (25-x)/4 )^{2}A = π( x

^{2}/ 4π^{2}) + ( (625 - 50x + x^{2}) / 16 )A = x

^{2}/4π + 1/16 ( x^{2}- 50x + 625 )A = x

^{2}/4π + x^{2}/16 - 50x/16 + 625/16Now let's differentiate:

dA/dx = x/2π + x/8 - 25/8

dA/dx = 4x/8π + πx/8π - 25/8

dA/dx = (4+π)/8π * x – 25/8

dA/dx = (4+π)/8π * x – 25/8

And now, we just have to set the derivative equal to 0 to find our maxima and/or minima.

0 = (4+π)/8π * x – 25/8

(4+π)/8π * x = 25/8

x = 25π/(4+π) = 10.99752.. ~ 11

We find that we have only one extrema, so let’s check to see if it is a maxima or minima. We will do so by checking areas around the critical point for the sign of the slope of their derivative. Looking at the equation for dA/dx, when dA/dx was 0, we found that:

0 = (4+π)/8π * x - 25/8 when x = 11

0 = (4+π)/8π * x – 25/8

(4+π)/8π * x = 25/8

x = 25π/(4+π) = 10.99752.. ~ 11

We find that we have only one extrema, so let’s check to see if it is a maxima or minima. We will do so by checking areas around the critical point for the sign of the slope of their derivative. Looking at the equation for dA/dx, when dA/dx was 0, we found that:

0 = (4+π)/8π * x - 25/8 when x = 11

So if x = 11, the first portion of the expression [ (4+π)/8π * x ] is equal to 25/8. If that's the case, then:

If x < 11, the first portion of the expression will be smaller than 25/8 and the derivative will be negative.

If x > 11, the first portion of the expression will be greater than 25/8 and the derivative will be positive.

If the derivative is negative as it approaches x = 11 from the left, 0 at x = 11, and positive as it departs x = 11 on the right, then x = 11 is a minima. This minima represents the point at which the surface area is smallest.

Our goal was to determine how we could cut the string to obtain a circle and square with the smallest combined surface area.

1. We built an equation which related the combined surface area to the length of the circumference of the circle (and indirectly to the length of the square as well).

2. We derived this equation to give us a formula for the rate of increase or decrease in surface area with respect to the length of rope we cut.

3. We set the derivative equal to 0 to obtain any maxima or minima and then determined that this point was a minima.

The rope should be cut so that the length of rope used to form the circle will be 11 inches long.

Was this the most efficient way to solve this problem? Have you run in to some other tricky optimization or related rates problems? Share your thoughts below!

Best of luck in your studies,

-Skyler

-Skyler

## Comments

^{2}+ s^{2 }subject to the constraint 2πr + 4s = 25.^{2}+ s^{2}) +λ (2πr + 4s), with Lagrange multiplier λ, finds its partial derivatives and set them equal to zero:_{r}= 2πr + 2π λ =0, L_{s}= 2s + 4λ =0.^{2}/π.