Share

# A Wire of Length 20 M is to Be Cut into Two Pieces. One of the Pieces Will Be Bent into Shape of a Square and the Other into Shape of an Equilateral Triangle. Where the We Should Be Cut So that the - Mathematics

#### Question

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

#### Solution

$\text {Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length x and y, respectively . Then},$

$x + y = 20 .......... \left( 1 \right)$

$\text { Perimeter of square }, 4\left( Side \right) = x$

$\Rightarrow \text { Side } = \frac{x}{4}$

$\text { Area of square = }\left( \frac{x}{4} \right)^2 = \frac{x^2}{16}$

$\text { Perimeter of triangle }, 3\left( \text { Side } \right) = y$

$\Rightarrow \text { Side } = \frac{y}{3}$

$\text { Area of triangle } = \frac{\sqrt{3}}{4} \times \left( \text { Side } \right)^2 = \frac{\sqrt{3}}{4} \times \left( \frac{y}{3} \right)^2 = \frac{\sqrt{3} y^2}{36}$

$\text { Now,}$

$z =\text { Area of square + Area of triangle }$

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

$\Rightarrow z = \frac{x^2}{16} + \frac{\sqrt{3} \left( 20 - x \right)^2}{36} .................\left[ \text { From eq } . \left( 1 \right) \right]$

$\Rightarrow \frac{dz}{dx} = \frac{2x}{16} - \frac{2\sqrt{3}\left( 20 - x \right)}{36}$

$\text { For maximum or minimum values of z, we must have }$

$\frac{dz}{dx} = 0$

$\Rightarrow \frac{2x}{16} - \frac{\sqrt{3}\left( 20 - x \right)}{18} = 0$

$\Rightarrow \frac{9x}{4} = \sqrt{3}\left( 20 - x \right)$

$\Rightarrow \frac{9x}{4} + x\sqrt{3} = 20\sqrt{3}$

$\Rightarrow x\left( \frac{9}{4} + \sqrt{3} \right) = 20\sqrt{3}$

$\Rightarrow x = \frac{20\sqrt{3}}{\left( \frac{9}{4} + \sqrt{3} \right)}$

$\Rightarrow x = \frac{80\sqrt{3}}{\left( 9 + 4\sqrt{3} \right)}$

$\Rightarrow y = 20 - \frac{80\sqrt{3}}{9 + 4\sqrt{3}} .............\left[ \text { From eq }. \left( 1 \right) \right]$

$\Rightarrow y = \frac{180}{9 + 4\sqrt{3}}$

$\frac{d^2 z}{d x^2} = \frac{1}{8} + \frac{\sqrt{3}}{18} > 0$

$\text { Thus, z is minimum when } x = \frac{80\sqrt{3}}{\left( 9 + 4\sqrt{3} \right)} \text { and }y = \frac{180}{9 + 4\sqrt{3}} .$

$\text { Hence, the wire of length 20 cm should be cut into two pieces of lengths }\frac{80\sqrt{3}}{\left( 9 + 4\sqrt{3} \right)} \text { m and } \frac{180}{9 + 4\sqrt{3}} m .$

#### Notes

The solution given in the book is incorrrect . The solution here is created according to the question given in the book.

Is there an error in this question or solution?