Question

A piece of wire 20 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is: (a) maximum? (b) minimum? 

Answer 1

i)

Let x meters be the side of square , so 4$x$ meters be the total amount of wires used for square .(with 0$\leq4x\leq20 \implies 0\leq x\leq5).$ For equilateral triangle 20-4$x$ , wire remains .and the side will be $\dfrac{20-4x}{3}$

Given the side l of an equilateral triangle , the height is , is using the pitagoras theorem , we get -$h=\dfrac{l{\sqrt3}}{2},$ so the area of triangle will become ,

$A=\dfrac{l}{2}\dfrac{l{\sqrt 3}}{2}\dfrac{1}{2}$

so the total area is -

$A=x^{2}+(\dfrac{20-4x}{3})^{2}\dfrac{\sqrt 3}{4}$

$A{'}=2x+\dfrac{\sqrt 3}{4}\dfrac{1}{9}.2(20-4x).(-4)=2x-\dfrac{2{\sqrt 3}}{9}(20-4x)$

$A^{'}>0 \ if$

$=2x-\dfrac{2{\sqrt3}}{9}(20-4x)>0$

$x>\dfrac{40\sqrt 3}{2(9+4{\sqrt 3})}$

$x>\dfrac{20(3{\sqrt 3}-4)}{11}=$

So the function area is decreasing in [0,$x)$ and growing in (x,5] so the point is $(\bar x,5]$ is minimum of the function . The conclusion is that maximum (absolute maximum) is minimum of side of the function of the interval of function $x$ ,0,5 .

if x=0 , then the area will be - $A=19.2$

if x=0 , then the area will be $x$ =25.

in case the maximum area will be there where all the wire used as square .

ii)

Let the two pieces be $x\ and\ y.$

Given,

$=x+y=20$

Perimeter of square =$x$

x

side$\ =\dfrac{x}{4}$

Area of square =$\dfrac{x^{2}}{16}$

Perimeter of triangle $=y$

Side$\ =\dfrac{y}{3}$

We know , that area of triangle is given by $=\dfrac{\sqrt3}{4}a^{2}$

$=\dfrac{\sqrt3}{4}(\dfrac{y}{3})^{2}=\dfrac{\sqrt{3}}{36}y^{2}$

$z=$ area of square +area of triangle

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

Now replacing the value of $y,$

Replacing $y=20-x,$

$z=\dfrac{x^{2}}{16}+\dfrac{\sqrt3}{36}(20-x)^{2}$

$=\dfrac{dz}{dx}=\dfrac{x}{8}-\dfrac{\sqrt3}{18}(20-x)$

equate ,$\dfrac{dz}{dx}=0$

we have ,

$=\dfrac{x}{8}-\dfrac{\sqrt 3(20-x)}{18}=0$

$=\dfrac{x}{8}=\dfrac{\sqrt 3(20-x)}{18}$

Solving the above equation , we get -

$x=\dfrac{80\sqrt 3}{9+4\sqrt 3}$

$y=20-\dfrac{80\sqrt 3}{9+4\sqrt 3}$

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

Thus , z is minimum when $x=$ $\dfrac{80\sqrt 3}{9+4\sqrt 3}$ $y=\dfrac{180}{9+4\sqrt 3}$ .