Answer to Question #124035 in Calculus for Nii Laryea

Question

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

x cm - on square

(100 - x) cm - on equilateral triangle

Then $S(x) = (\dfrac{x}{4})^2 + (\dfrac{100-x}{3})^2 \cdot sin60^\circ \cdot 0,5 =$ $\dfrac{x^2}{16} +\dfrac{(100-x)^2}{12 \sqrt{3}}$

$S'(x) = \dfrac{x}{8} + \dfrac{2(100-x)}{12\sqrt{3}} \cdot (-1) = x (\dfrac{1}{8}+\dfrac{1}{6\sqrt{3}}) -$ $\dfrac{100}{6\sqrt{3}}$

When $S'(x) = 0$ it changes the sign from "-" to "+", so the sum of the areas is minimized (if found x is between 0 and 100)

$x (\dfrac{6\sqrt{3}}{8} + 1) = 100;$

$x =\dfrac{100}{ \dfrac{6\sqrt{3}}{8} + 1}$ - it is between 0 and 100. So, The first piece (for square) have the lenght $(\dfrac{100}{ \dfrac{6\sqrt{3}}{8} + 1} ) cm$ and the second piece (for equilateral triangle) have the lenght $(100 - \dfrac{100}{ \dfrac{6\sqrt{3}}{8} + 1} ) cm$