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QJEEM21Maae50: JEE Main 2021 Mathematics - Class 12 JEEM21Maae50 Solution: A wire of length 20 m is to be cut into two pieces. One of the…

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 :

Multiple Choice Options

A ${5 \over {2 + \sqrt 3 }}$
B ${{10} \over {2 + 3\sqrt 3 }}$
C ${5 \over {3 + \sqrt 3 }}$
D ${{10} \over {3 + 2\sqrt 3 }}$

Step-by-Step Verified Solution

Let the wire is cut into two pieces of length x and 20

-

x.

JEE Main 2021 (Online) 27th August Morning Shift Mathematics - Application of Derivatives Question 91 English Explanation

Area of square =

{\left( {{x \over 4}} \right)^2}

Area of regular hexagon =

6 \times {{\sqrt 3 } \over 4}{\left( {{{20 - x} \over 6}} \right)^2}

Total area =

A(x) = {{{x^2}} \over {16}} + {{3\sqrt 3 } \over 2}{{{{(20 - x)}^2}} \over {36}}

A'(x) = {{2x} \over {16}} + {{3\sqrt 3 \times 2} \over {2 \times 36}}(20 - x)( - 1)

A'(x) = 0 at

x = {{40\sqrt 3 } \over {3 + 2\sqrt 3 }}

Length of side of regular Hexagon

= {1 \over 6}(20 - x)

= {1 \over 6}\left( {20 - {{4.\sqrt 3 } \over {3 + 2\sqrt 3 }}} \right)

= {{10} \over {2 + 2\sqrt 3 }}

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