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Show that the rectangle of largest possible area, for a given perimeter, is a square.

My Attempt:

Let $\textrm {length}=x$ and $\textrm {breadth}=y$. Then,

Perimeter of rectangle $=2(x+y)$

Also, Area of rectangle $=x.y$ $$A=x.y$$ Now, $\dfrac {dA}{dx}=x.\dfrac {dy}{dx}+y.\dfrac {dx}{dx}$ $$=x.\dfrac {dy}{dx}+y$$

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You need to use the perimeter of the rectangle $P=2(x+y)$, or $y=P/2-x$. It should be easy to take the derivative. Replace $y$ and $dy/dx$ in your last equation, equate it to $0$, and you obtain that $x=P/4$. Then $y=P/4=x$.

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Here's a shorter solution without using : By the AM-GM inequality, $\displaystyle \sqrt{A} = \sqrt{xy} \le \frac{x+y}{2} \iff A \le \frac{(x+y)^2}{4},$ equality holds iff $x = y$.

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You want to solve this problem $$\max_{x+y =a, \, x \ge 0, \, y \ge 0} xy$$ which is equivalent to ($y = a-x$)$$\max_{0 \le x \le a} x(a-x)$$ After differentiation $x= a/2$ and so $y= a-x = a/2 = x$

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