a wire of length (L) is divided into 2 parts. One forms a circle and the other forms a square. How should the wire be cut so the total area is a minimum?

Question

what are the steps to do this?

the answer is x= 4L / 4+pi

Answer 1

Square's Perimeter = 4s

Circle's Circumference = 2pi(r)= L-2s

Circle's radius = r = (L-4s)/2pi

Circle's Area = pi(r^2) = = pi[L-4s)/2pi]^2

Square's Are = s^2

Total Area = A = s^2 + (1/4pi)(L^2 -8Ls + 16s^2)

take the derivative and set equal to zero

A'(s) = 2s + (1/4pi)(-8L+32s) = 0

2s +(8s-2L)/pi = 0

s +(4s-L)/pi = 0

s(1+4/pi) = L/pi

s = (L/pi)/(4+pi)/pi = L/(4+pi)

4s = 4L/(4+pi) = part used to form a square

L-4s = L-4L/(4+pi) = piL/(4+pi) = part used to form a circle

Answer 2

Let's think a bit before diving in.

For each shape, we're going to have some fixed length of wire. That wire is going to completely determine the shape, because each shape is determined by a single parameter (the side length s for the square, the radius r for the circle). So, we're going to want to get s and r, and then use them to get the areas of the two shapes. That done, we can just add them to get the total area enclosed. From there, it's just our usual calculus operations.

So let's do it. Let's say that of the wire L, we allocate x to the square, leaving L-x for the circle.

We'll focus on the square first. If we have x length of wire, that means that the perimeter of the square must be x (because that's what the perimeter means, the length enclosing the area!). We know that the perimeter of a square in general is 4s, so we can find s

4s=x

s=x/4

That done, we can get the area of the square. It's just s²!

A_S=s²

=(x/4)²

=x²/16

Now that we've got the area of the square, let's do the same for the circle. The circle gets L-x wire, which is again the perimeter. The perimeter for a circle is 2 pi r, so we can get the radius:

2 pi r=L-x

r=(L-x)/(2 pi)

The area of a circle in terms of its radius is pi r², so the area of the circle in terms of the allocated wire is

A_C=pi r²

=pi [(L-x)/(2 pi)]²

=[1/(4 pi)] x² - [L/(2 pi)] x + L²/(4 pi)

(I skipped some simplifying algebra). Now that we've got the areas of the both the square and circle, we add these together to get the total area

A=A_S+A_C

=[x²/16]+[1/(4 pi) x²-L/(2 pi) x+ L²/(4 pi)]

=[1/(4 pi)+1/16] x² - [L/(2 pi)] x + L²/(4 pi)

We now have A as a function of x. Time for calculus! Take the derivative and set to 0 to find the local extrema

dA/dx=2[1/(4 pi)+1/16] x - L/(2 pi)=0 -> x=4L/(4+pi)

We need to verify that this is actually a minima. If we take the second derivative, we find

d²A/dx²=2[1/(4 pi)+1/16]

Since this is always positive no matter the value of x, it's certainly positive at x=4L/(4+pi). This confirms that we have a minimum.

So, we want to use 4L/(4+pi) for the square, and L-4L/(4+pi)=pi L/(4+pi) for the circle.

Answer 3

I will show you how I did this

C=circumference of circle and S=perimeter of square and A is area

C+S=L, i.e. S=L-C

A=(C/2π)²π + (S/4)² = [C²/4π] + (L²-2LC+C²)/16

Differentiate and set derivative to 0:

8C - 2πL+2πC = 0

C=πL/(4+π) and S = 4L/(4+π)

Answer 4

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