Question

Show that of all the rectangles with a given perimeter, the square has the largest area.

Answer 1

Hint: We have to prove the given statement by the area and perimeter of a rectangle and to prove the maxima by applying the maxima and minima conditions.
From the given, we have to show that the rectangle with a given perimeter implies that the square and also the area of the rectangle be large when the rectangle is square.

Formula used: Let the length of the rectangle be \[{\text{x cm}}\]and the breadth of the rectangle be\[{\text{y cm}}\].
Let the area of the rectangle denoted by \[{\text{A}}\] and the perimeter of the rectangle be \[{\text{P}}\].
Area of the Rectangle, \[{\text{A}} = {\text{xy}}\].
Perimeter of the Rectangle, \[{\text{P}} = 2\left( {{\text{x + y}}} \right)\].

Complete step-by-step answer:
We know that, the perimeter of the rectangle, \[{\text{P}} = 2\left( {{\text{x + y}}} \right)\].
\[ \Rightarrow {\text{P = 2x + 2y}}\]
We are going to solve this for \[{\text{y}}\],
\[ \Rightarrow {\text{P - 2x = 2y}}\]
Rearranging the terms we get,
\[ \Rightarrow {\text{y = }}\dfrac{{{\text{P - 2x}}}}{{\text{2}}}\]
We know that the area of the rectangle, \[{\text{A = \;xy}}\].
Now, we have to substitute \[{\text{y}}\] term in the above area formula, A.
$ \Rightarrow {\text{A = x}}\left( {\dfrac{{{\text{P - 2x}}}}{{\text{2}}}} \right)$
Now, multiply the ${\text{x}}$ into the numerator, then we get
$ \Rightarrow {\text{A = }}\dfrac{{{\text{Px - 2}}{{\text{x}}^{\text{2}}}}}{{\text{2}}}$
On differentiating both sides, with respect to \[{\text{x}}\], we get
\[ \Rightarrow \dfrac{{{\text{dA}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{{\text{Px - 2}}{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right)\]
\[ \Rightarrow \dfrac{{{\text{dA}}}}{{{\text{dx}}}} = \dfrac{{{\text{P - 4x}}}}{2}\]
For maxima and minima, put \[\dfrac{{{\text{dA}}}}{{{\text{dx}}}} = 0\].
$ \Rightarrow \dfrac{{{\text{P - 4x}}}}{{\text{2}}}{\text{ = 0}}$
Simplifying we get,
$ \Rightarrow {\text{P - 4x = 0}}$
Hence,
$ \Rightarrow {\text{P = 4x}}$
From the perimeter of the rectangle, ${\text{P = 2x + 2y}}$ .
$ \Rightarrow 2{\text{x + 2y = 4x}}$
Simplifying we get,
$ \Rightarrow {\text{2y = 4x - 2x}}$
Subtracting the terms,
$ \Rightarrow {\text{2y = 2x}}$
Cancelling the common term $2$ on both sides,
$ \Rightarrow {\text{y = x}}$
Hence we get,
$ \Rightarrow {\text{x = y}}$
Therefore, the length and breadth of the rectangle with a given perimeter will be equal.
This implies that the sides of the rectangle are equal and it seems like the condition of the square. Since, the square has equal sides.
$\therefore $ Thus, the rectangle is square.
Now, we have to prove that the area of the rectangle is maximum. For the proof, we have to find the second derivative of ${\text{A}}$, and it be less than zero.
$\dfrac{{{{\text{d}}^{\text{2}}}{\text{A}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{{\text{dA}}}}{{{\text{dx}}}}} \right)$
Here we have to substitute $\dfrac{{{\text{dA}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{P - 4x}}}}{{\text{2}}}$ , we get
$\Rightarrow\dfrac{{{{\text{d}}^{\text{2}}}{\text{A}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{{\text{P - 4x}}}}{{\text{2}}}} \right)$
$\Rightarrow\dfrac{{{{\text{d}}^{\text{2}}}{\text{A}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{\text{P}}}{{\text{2}}}} \right) - \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{{\text{4x}}}}{{\text{2}}}} \right)$
Here, on differentiating the constant always be zero. Then, we get
\[\Rightarrow\dfrac{{{{\text{d}}^{\text{2}}}{\text{A}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}0 - 2\]
\[\Rightarrow\dfrac{{{{\text{d}}^{\text{2}}}{\text{A}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }} - 2\]
Therefore, \[\dfrac{{{{\text{d}}^{\text{2}}}{\text{A}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}} < 0\] which shows that the area is maximum.

Hence, the area is maximum when the rectangle is a square.
Show that of all the rectangles with a given perimeter, the square has the largest area.

Note: Square and rectangle has 4 sides each. All the sides of a square are equal and opposite sides of the rectangle are equal. Internal angles of both square and rectangle measures to be 90. Each of them have 2 diagonals each.