A) Parallelogram
B) Trapezium
C) Square
D) None of these
Correct Answer: C
Solution :
We know that perimeter of a rectangle \[S=2(x+y)\], where x and y are adjacent sides Þ \[y=\frac{S-2x}{2}\]. Now area of rectangle, \[A=xy=\frac{x}{2}(S-2x)=\frac{1}{2}(Sx-2{{x}^{2}})\] Differentiating w.r.t. x of A, we get \[\frac{dA}{dx}=\frac{1}{2}(S-4x)=0\,\,\,\therefore x=\frac{S}{4}\]and \[y=\frac{S}{4}\] Again \[\frac{{{d}^{2}}A}{d{{x}^{2}}}=-ve\] Hence the area of rectangle will be maximum when rectangle is a square.
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