Question

A wire of length 20 cm is bent in the form of square find the area of a square so far​

Answer 1

Rectified question:-

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

Solution:-

Let,

$\red\diamond$the 2 pieces are x,y

Given,

$\red\diamond$x + y =20

$\red\diamond$The Perimeter of square = x

$\red\diamond$The Side = $\sf{\frac{x}{4}}$

$\red\diamond$The Area of square = $\sf{\frac{x^2}{16}}$

$\red\diamond$The Perimeter of Triangle = y

$\red\diamond$The Side = $\sf{\frac{y}{3}}$

$\mapsto$ Area of Triangle = $\sf{\frac{\sqrt{3}}{4}\times(\frac{y}{3})^{3}}\implies{\bf{\frac{\sqrt{3}y^{2} }{36}}}$

Here, z = area of square + area of the triangle

     $\mapsto\sf{z=\frac{x^{2} }{16}+\frac{\sqrt{3}y^{2} }{36}}$

     $\mapsto\sf{z=\frac{x^{2} }{16}+\frac{\sqrt{3}(20-x)^{2} }{36}}$

By differentiating the equation,

$\implies\sf{\frac{dy}{dx} =\frac{x}{8}-\frac{\sqrt{3}(20-x) }{18}}$

We know, $\sf{\frac{dy}{dx}=0}$

Now we have,

$\implies\sf{\frac{x}{8}-\frac{\sqrt{3}(20-x)}{18}}$

$\implies\sf{\frac{x}{8}=\frac{\sqrt{3}(20-x)}{18}}$

$\dashrightarrow{\boxed{\bf{x=\frac{80-\sqrt{3}}{9+4\sqrt{3}}}}}$

$\dashrightarrow{\boxed{\bf{y=\frac{180}{9+4\sqrt{3}}}}}$

$\sf{\frac{d^{2}y }{dx^{2}}=\frac{1}{8}+\frac{\sqrt{3} }{18}>0}$

Thus z is minimum, when $\sf{x=\frac{80-\sqrt{3}}{9+4\sqrt{3}}}$ and $\sf{y=\frac{180}{9+4\sqrt{3}}}$

∴ The wire of length 20cm should be cut into pieces of lengths $\bf{x=\frac{80-\sqrt{3}}{9+4\sqrt{3}}}$ and $\bf{y=\frac{180}{9+4\sqrt{3}}}$.

Hope it helps!!