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

Question

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

TestHub · Accepted Answer

Let the wire is cut into 2 pieces of length $y & 20 - y$

Therefore side of square will be $= \frac{y}{4}$ and hexagon will be $= \frac{20 - y}{6}$

Area of square $= {(\frac{y}{4})}^{2}$

Area of hexagon $= 6 \times \frac{\sqrt{3}}{4} {(\frac{20 - y}{6})}^{2}$

Total area $= \frac{y^{2}}{16} + \frac{\sqrt{3} {(20 - y)}^{2}}{24}$

On differentiating,

$A^{'} (y) = \frac{2 y}{16} + \frac{2 \sqrt{3} (20 - y) (- 1)}{24}$

$A^{'} (y) = 0 a t y = \frac{40 \sqrt{3}}{3 + 2 \sqrt{3}}$

Length of side of regular hexagon $= \frac{1}{6} (20 - y) = \frac{1}{6} (20 - \frac{40 \sqrt{3}}{3 + 2 \sqrt{3}}) = \frac{10}{3 + 2 \sqrt{3}}$

Mathematics - Application of Derivative Question with Solution | TestHub

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