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Question

I'm really confused
A 48 inch piece of string is cut into 2 pieces. One piece is used to form a circle, while the other is used to form a equilateral triangle. How should the string be cut so that the sum of the area is a minimum.
Thanks for any help, I tried working it out for like an hour and can't get it right :S

Answer 1

Cut the string into 2 lengths, a and b.

So we have to start with

a+b=48

Assume a makes the circle. The length of the string is the circumference so

a = 2πr

The area of the circle is πr²

a = 2πr --> πr=(a/2) and r=a/(2π)

πr² = (a/2) a/(2π) = a²/(4π)

We have a piece of string to make an equilateral triangle. Since the string is of length b, each side must be of length b/3.

A well known result for the area of an equilateral triangle with sides of length s is

Area=¼s²√3

So the area of our triangle is

Area=¼(b/3)²√3
Area=b²√3/36

So the combined area of our circle and triangle is

A = a²/(4π) + b²√3/36

Since a+b=48 we can substitute b as b = 48-a

A = a²/(4π) + (48-a)²√3/36

So let's now take the derivative of A with respect to a

dA/da = a/(2π) - (48-a)√3/18

We set the derivative to 0 to find the critical points

0 = a/(2π) - (48-a)√3/18
a/(2π) = (48-a)√3/18
a = (48-a)(2π)√3/18
a = 16π√3/3 - aπ√3/9
a + aπ√3/9 = 16π√3/3
a(9 + π√3)/9 = 16π√3/3
a = 48π√3/(9 + π√3)

If this is a minimum then d²A/da² > 0

dA/da = a/(2π) - (48-a)√3/18
d²A/da² = 1/(2π) + √3/18 > 0

So it is a minimum and the minimum area is when

a = 48π√3/(9 + π√3) (which is roughly 18.1)

b = 48-a --> I'll leave this to you.

Answer 2

This is a minimisation problem subject to a condition, let's have a go at working out the conditions and solving.

Firstly let's go right ahead and cut that string up into x (to form a circle) and y (to form the triangle).

An area of a circle is Pi * r^2, the circumference is 2 * Pi * r [where r is the radius] now we know the circumference has length x so we can say;

x = 2*Pi*r => r = x/(2*Pi)
giving area C=Pi*[x/(2*Pi)]^2

simplifying gives C=x^2/(4*Pi)

similarly for the triangle we have to calculate the area of an equilateral triangle each side length y/3. Using Pythagoras we can see the hypotenuse is y/3 and one edge is y/6 we thus have

height = sqrt [ (y/3)^2 - (y/6)^2 ] = sqrt [3*y^2 / 36] = y/2*sqrt(3)
and
area T = 1/2 * base * height
=1/2 * y/3 * y/2*sqrt(3)
=y^2/[12*sqrt(3)]

So our aim is to minimise C+T = x^2/(4*Pi) + y^2/[12*sqrt(3)] subject to x+y=48.

This conditional problem can be solved using legrangian multipliers

L= x^2/(4*Pi) + y^2/[12*sqrt(3)] + j*[48-(x+y)]

Where j is the legrangian multiplier, differentiating WRT x, y and j and setting equal to 0 gives;

Lx = x/(2*Pi) - j = 0 (1)
Ly = y/6*sqrt(3) - j = 0 (2)
Lj = 48 - (x+y) = 0 (3)

solving (1) and (2) for j gives
j = x/(2*Pi)
j = y/6*sqrt(3)

=> x/(2*pi) = y/6*sqrt(3)
=> x = y*Pi/3*sqrt(3)

substituting into (3)

y*Pi/3*sqrt(3) + y = 48
=> y * [1+ Pi/3*sqrt(3)] = 48
=> y = 48/[1+ Pi/3*sqrt(3)] = 29.91
=> x=48-y=18.09.

There may be an easier way but hopefully this isn't too hard to follow.

Answer 3

u cut the string into two peices,let one is=x, and the other is =48-x
for the circle,perimeter=x
2*pi*R =x. and for the triangle the perimeter is = 48-x
48 -x = 4*S (side length)
sum of areas = pi *R^2 + S^2
SOA = pi * (x / 2pi)^2 + ((48 - x)/4)^2
SOA = (1/4pi) x^2 + (12 - .25 x)^2
SOA= (1/4pi) x^2 + 144 - 6x + .0625 x^2
now diffrentite wrt x,
S' = (1/2pi) x - 6 + .125 x
min. area occur at S' =0
0.159 x +0.125 x = 6
0.284 x = 6
x = 21.1 inch
so you cut the string into a part of length 21.1 inch to form the circle,and a part of length 26.9 to form the triangle.

Answer 4

Let x = one piece of string
Then 48-x = length of other piece of string

Use x to form an equilateral triangle
Area of triangle = x^2sqrt(3)/36

Use 48-x to form circle
C= 48-x = 2pir
So A = (48-x)^2/(4pi)

Total area = T = x^2sqrt(3)/36 + (48-x)^2/(4pi)

Now find dT/dx, set it = 0, and solve for x

Answer 5

The total length of the string is equal to the sum of perimeters of the circle and triangle

let r= radius of circle and s = side of triangle

Total Perimeter = Perimeter of (circle + triangle)

2pi(r) + 3s = 48

3s = 48 - 2pir

3s = 2(24 - pi r)

s = 2/3(24 - pi r)

total area A = area of (circle + triangle)

A = pir^2 + sqrt(3)s^2/4

A = pir^2 +(1/4)(4/9)sqrt(3)[(24 - pi r]^2

A = pir^2 +(1/9)sqrt(3)[(24 - pi r]^2


so r lies in the region 0 and 24/pi

now find out derivative of A

A'(r) = 2pir + sqrt(3)/9*[(2)(24 -pir)*(-pi)]

A'(r) = 2pir - (2sqrt(3)pi /9)(24 -pi r)

solve A'(r) = 0

2pir = (2sqrt(3)pi /9)(24 -pi r)

r = 24 sqrt(3)/9 - pi r sqrt(3)/9

9r = 24sqrt(3) - pir sqrt(3)

r(9 + pi sqrt(3)) = 24sqrt(3)

r = 24sqrt(3)/(9 + pisqrt(3)) = 2.88

Now find out the A at 0, 24/pi and 2.88

A = pir^2 +(1/9)sqrt(3)[(24 - pi r]^2

A(0) = (sqrt(3)/9)*(24)^2 = 110.85

A(24/pi) = pi(24/pi)^2 +sqrt(3)/9[24 - pi24/pi]^2

=>(24)^2/pi + 0 = 183.35

A(2.88) = pi(2.88)^2 + sqrt(3)/9[24 - pi(2.88)]^2

=>26.06 + 0.192(223.56)

=>69

By comparing the values when r = 2.88 total area is minimum

2pir = 2(pi)(2.88) = 18.1

3s = 48 - 18.1 = 29.9

So the string is to be cut into two pieces of length 18.1 in for circle and 29.9 in for triangle