A piece of string with a length of n cm is cut into 2 pieces. One of the pieces will be used to make a square, while the other piece will be used to...

Question

A piece of string with a length of n cm is cut into 2 pieces. One of the pieces will be used to make a square, while the other piece will be used to make an equilateral triangle. How should the string be cut in order to minimize the combined area of the square and triangle?

 

Can you please include the process to solve this problem?

Answer 1

Calculus Optimization !!

 

We are trying to minimize the amount of area between the sum of 2 geometric figures

 

We will model the area formula in one variable and then apply calculus

 

-------------But first, let's look at our constraint, the length of the string----------------

 

n is the length of the string

 

We will use x amount of it for the triangle

 

that leaves us n-x of it for the square

 

each of these amounts x, and n - x are the perimeters of our figures

 

Square's Perimeter = 4*side

 

Traiangle's perimeter would be 3*side

 

x/3 is the side length of the triangle

 

(n-x)/4 is the side length of the square

 

___________________________________________________________________________________________

 

 

OK now for the sum of our 2 areas:

 

square area plus triangle area

 

The area of the sqaure is the square of the side length:  ((n-x)/4)^2  = (n-x)^2/16

 

The area of the triangle is a little trickier. 

 

The base will be x/3

 

There are many different types of triangles named by their side length, and angle measure. An equilateral triangle has the following properties.

Equilateral triangles have three equal side lengths
Equilateral triangles also have three equal angles of 60 degrees
They also have three equal altitudes
Is the height of an equilateral triangle equal to its side length. No It is easy to imagine with everything being equal that the height, and the side length would be equal, but they are not.
The height is equal to side/2 x √3

 

Our side length is x/3  so our height is  sqrt(3)x/6

 

Therefore the area would 1/2(x/3)sqrt(3)x/6   = 1/2base*height

 

Sum of area of 2 figures is A(x) = (n-x)^2/16  +  1/2(x/3)sqrt(3)x/6 

 

We can clean this up to be A(x) = (n-x)^2/16 + sqrt(3)x^2/36

 

Now we use calculus ---------------------------------------------------------------------

 

We will derive our function

 

A '(x) = -(n-x)/8 + sqrt(3)x/18

 

Set A '(x) = 0 and clean up the right side

 

0 = -8n+8x + sqrt(3)x/8  multiply both sides by 8

 

0 = -64n+64x +sqrt(3)x = -64n +(64+sqrt(3))x    bring 64n over to the left side

 

64n = (64+sqrt(3))x          Now divide both sides by (64+sqrt(3))

 

64n/(64+sqrt(3)) = x