JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 11)
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 :
$${5 \over {2 + \sqrt 3 }}$$
$${{10} \over {2 + 3\sqrt 3 }}$$
$${5 \over {3 + \sqrt 3 }}$$
$${{10} \over {3 + 2\sqrt 3 }}$$
Explanation
Let the wire is cut into two pieces of length x and 20 $$-$$ x.
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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 $$ = {1 \over 6}(20 - x)$$
$$ = {1 \over 6}\left( {20 - {{4.\sqrt 3 } \over {3 + 2\sqrt 3 }}} \right)$$
$$ = {{10} \over {2 + 2\sqrt 3 }}$$
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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