A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is :
Show Hint
For area optimization with fixed perimeter, the ratio of side lengths often involves the ratios of their geometric constants. Differentiating the area function is the most robust method.
Step 1: Understanding the Concept:
This is a Maxima/Minima problem. Total Area = Area(Square) + Area(Hexagon). We need to minimize this subject to the length constraint. Step 2: Detailed Explanation:
Let side of square be $s$ and side of hexagon be $h$.
Total length = $4s + 6h = 20 \implies 2s + 3h = 10 \implies s = \frac{10 - 3h}{2}$.
Area $A = s^2 + \frac{3\sqrt{3}}{2} h^2$.
\[ A(h) = \frac{(10 - 3h)^2}{4} + \frac{3\sqrt{3}}{2} h^2 \]
For minimum area, $dA/dh = 0$:
\[ \frac{2(10-3h)(-3)}{4} + 3\sqrt{3} h = 0 \]
\[ -\frac{3}{2} (10 - 3h) + 3\sqrt{3} h = 0 \]
Divide by 3:
\[ -5 + 1.5h + \sqrt{3} h = 0 \implies h(1.5 + \sqrt{3}) = 5 \]
\[ h(\frac{3 + 2\sqrt{3}}{2}) = 5 \implies h = \frac{10}{3 + 2\sqrt{3}} \] Step 3: Final Answer:
Side of the hexagon is $\frac{10}{3 + 2\sqrt{3}}$.
