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 first piece have a length of 4x and the second piece have the length of (100-4x). Therefore, the area of square is $x^2$. The length of edge of triangle is $a=\dfrac{100-4x}{3}$ and the area is $\dfrac{\sqrt{3}}{4} a^2 =\dfrac{\sqrt{3}}{4} \cdot\dfrac{(100-4x)^2}{9} .$

The sum of areas is $S(x) = x^2 + {\sqrt{3}} \cdot\dfrac{(100-4x)^2}{36} .$

Next, we calculate the derivative of S(x):

$S'(x) = 2x + \dfrac{\sqrt{3}}{36}\cdot2\cdot(100-4x)\cdot(-4) = x\left(2+\dfrac{8\sqrt{3}}{9}\right) - \dfrac{200\sqrt{3}}{9}.$

We assume $S'(x) = 0$ and get $x = \dfrac{100\sqrt3}{9+4\sqrt3} \approx 10.874.$

Therefore, the first piece has length of $\dfrac{400\sqrt3}{9+4\sqrt3} \approx 43.496$ cm, the second piece has length of $\dfrac{900}{9+4\sqrt3} \approx 56.504$ cm.