Question

A wire of length 20cm is to be cut into two pieces. One of the pieces will bend into the shape of a square and another into an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?

Answer 1

Initially we will consider 2 variables for the cut pieces. With the help of it, we will find the sides of the shapes formed, Which will help us to find the area of the two shapes formed. And finally, add both the areas also we will differentiate the equation formed by adding the area of a square and equilateral triangle which will be used to find the minimum points using the following fact that a function $f(x)$ is said to be minimal if the value $\dfrac{{{d^2}f}}{{d{x^2}}} > 0$ and the point the function will be minimum is $\dfrac{{df}}{{dx}} = 0$.

Complete step by step solution:
It is given that, the length of the wire is $20cm$. the wire cut into two pieces such that, one of the pieces will bend into the shape of a square and another into an equilateral triangle.
Let us consider the length of the two pieces are $x\& y.$
According to the problem $x + y = 20$.
Now, the length $x$ is bent into the square. So, the perimeter of the square is $x$.
Perimeter of a square is $x$ implies $4a = x$, where $a$ is the side of the square.
Each side of the square is $\dfrac{x}{4}$cm.
Also, we know that area of a square is ${a^2}$ where a is the side of the square,
Then the area of the square is $\dfrac{{{x^2}}}{{16}}$
Again, the length $y$ is bent into the equilateral triangle. So, the perimeter of the equilateral triangle is $y$.
Perimeter of an equilateral triangle is $y$ implies $3b = x$, where $b$ is the side of the equilateral triangle.
Each side of the equilateral triangle is $\dfrac{y}{3}$cm.
Also, we know that area of an equilateral triangle is $\dfrac{{\sqrt 3 }}{4}{(b)^2}$.
Therefore area of the equilateral triangle $\dfrac{{\sqrt 3 }}{4}{(\dfrac{y}{3})^2} = \dfrac{{\sqrt 3 {y^2}}}{{36}}$
Now,
$z =$Area of the square + area of the equilateral triangle
$z = \dfrac{{{x^2}}}{{16}} + \dfrac{{\sqrt 3 {y^2}}}{{36}}$
Substitute the value $y = 20 - x$ in the above equation we get,
$z = \dfrac{{{x^2}}}{{16}} + \dfrac{{\sqrt 3 {{(20 - x)}^2}}}{{36}}$
Let us differentiate the above equation with respect to $x$ we get,
$\dfrac{{dz}}{{dx}} = \dfrac{{2x}}{{16}} - \dfrac{{\sqrt 3 \times 2(20 - x)}}{{36}}$
On further simplification we get,
$\dfrac{{dz}}{{dx}} = \dfrac{x}{8} - \dfrac{{\sqrt 3 (20 - x)}}{{18}}$… (1)
Equate, $\dfrac{{dz}}{{dx}} = 0$to find the minimum point we get,
$\dfrac{{dz}}{{dx}} = \dfrac{x}{8} - \dfrac{{\sqrt 3 (20 - x)}}{{18}} = 0$
$\dfrac{x}{8} = \dfrac{{\sqrt 3 (20 - x)}}{{18}}$
On cross multiplication and further calculation we get,
$x = \dfrac{{80\sqrt 3 }}{{9 + 4\sqrt 3 }}$
Let us substitute $x$ in $y = 20 - x$ then we get,
$y = \dfrac{{180}}{{9 + 4\sqrt 3 }}$
Now, differentiate equation (1) again we get,
$\dfrac{{{d^2}z}}{{d{x^2}}} = \dfrac{1}{8} + \dfrac{{\sqrt 3 }}{{18}} > 0$
Since$\dfrac{{{d^2}z}}{{d{x^2}}} > 0$, $z$ is minimum when, $x = \dfrac{{80\sqrt 3 }}{{9 + 4\sqrt 3 }};y = \dfrac{{180}}{{9 + 4\sqrt 3 }}$

$\therefore$ The length of the two pieces are $x = \dfrac{{80\sqrt 3 }}{{9 + 4\sqrt 3 }};y = \dfrac{{180}}{{9 + 4\sqrt 3 }}$.

Note:
We know that, if each side of a square is$a$, then its area is ${a^2}$ and its perimeter is $4a$.
If each side of an equilateral triangle is $b$, then its area is $\dfrac{{\sqrt 3 }}{4}{b^2}$ and its perimeter is $3b$. We should be careful while finding the value of the sides of the shapes which will lead to the required answer.