Question

b) A wire of length 100 centimeters is cut into two pieces. One piece is bent to form a square. The other piece is bent to form an equilateral triangle. Find the dimensions of the two pieces of wire so that the sum of the areas of the square and the triangle is minimized.

Answer 1

Let the length of sides of equilateral triangle is x cm each and sides of square is y cm each.

Sum of perimeters of triangle and square is 3x + 4y cm

According to question

3x+4y = 100

=> y = $\frac {100-3x}{4}$

Now area of the equilateral triangle is $\frac {√3x²}{4}$ and area of square is y² = $(\frac{100-3x}{4})^2$ = $\frac{(100-3x)^2}{16}$

Sum of the areas ,

A = $\frac {√3x²}{4}$ + $\frac{(100-3x)^2}{16}$

$\frac {dA}{dx}$ = $\frac {2√3x}{4}$ + $\frac{-6(100-3x)}{16}$

$\frac {dA}{dx}$ = $\frac {√3x}{2}$ - $\frac{3(100-3x)}{8}$

$\frac {d^2A}{dx^2}$ = $\frac {√3}{2} + \frac{9}{8} > 0$

For extreme values of A, $\frac {dA}{dx}=0$

=> $\frac {√3x}{2}$ - $\frac{3(100-3x)}{8}$ = 0

=> $\frac{(4√3+9)}{8}x = \frac {300}{8}$

=> x = $\frac {300}{4√3+9} = 18.83$

Since $\frac {d^2A}{dx^2} > 0$ , A is minimum at x = $\frac {300}{4√3+9} = 18.83$

and then y = $\frac {100-3*18.83}{4} = 10.88$

3x = $\frac {900}{4√3+9}$ , 4y = $\frac {400√3}{4√3+9}$

So actual dimensions of the two pieces of wire are $\frac {900}{4√3+9} cm$ and $\frac {400√3}{4√3+9} cm$

Approximately dimensions of the two pieces of wire are 56.50 cm and 43.50 cm