Question

How to show that of all the rectangles with a given perimeter, the one with the greatest area is a square?

Answer 1

Hint : Given problem tests the concepts of derivatives and their applications. These type questions can be easily solved if we keep in mind the concepts of maxima and minima. In the problem, we are required to prove that of all the rectangles with a given perimeter, the one with the greatest area is a square.

Complete step by step solution:
Let us consider the rectangles with perimeter as P.
Then, $P = 2\left( {l + b} \right)$ where l represents length of the rectangle and b represents breadth of the rectangle.
Also, Area of rectangle $= A = l \times b$
Now, we have to show that of all such rectangles, the greatest area is of a square. But for proving this fact, we have to first find the area function in one variable so that we can apply the maxima and minima concepts and first and second derivative tests.
So, $l + b = \left( {\dfrac{P}{2}} \right)$
$\Rightarrow l = \left( {\dfrac{P}{2} - b} \right)$
So, we can substitute this expression for length into the area function to obtain it in one variable solely.
$A = \left( {\dfrac{P}{2} - b} \right) \times b$
$\Rightarrow A = \dfrac{{Pb}}{2} - {b^2}$
Now, to find the maximum value of an area, we maximize the area function using the maxima and minima concepts.
Finding the first derivative of the area function with respect to b for the first derivative test, we get,
$\dfrac{{dA}}{{db}} = \dfrac{d}{{db}}\left( {\dfrac{{Pb}}{2} - {b^2}} \right)$
$\Rightarrow \dfrac{{dA}}{{db}} = \dfrac{d}{{db}}\left( {\dfrac{{Pb}}{2}} \right) - \dfrac{d}{{db}}\left( {{b^2}} \right)$
$\Rightarrow \dfrac{{dA}}{{db}} = \dfrac{P}{2} - 2b$
Equating $\dfrac{{dA}}{{db}}$ with zero so as to find the critical points.
$\Rightarrow \dfrac{{dA}}{{db}} = \dfrac{P}{2} - 2b = 0$
$\Rightarrow \dfrac{P}{2} = 2b$
Now, substituting the value of P from the equation $P = 2\left( {l + b} \right)$ , we get,
$\Rightarrow \dfrac{{2\left( {l + b} \right)}}{2} = 2b$
$\Rightarrow l + b = 2b$
$\Rightarrow l = b$
So, we get $l = b$ as a condition for a critical point.
Finding second derivative of area function with respect to b for the second derivative test, we get,
$\dfrac{{{d^2}A}}{{d{b^2}}} = \dfrac{d}{{db}}\left( {\dfrac{P}{2} - 2b} \right) = - 2$
Since the second derivative is negative, the critical point represents maxima. Hence, the greatest area occurs for $l = b$ . We know that a rectangle having equal length and breadth is a square.
So, we can conclude that of all rectangles with a given perimeter, a square has the greatest area.
Hence, Proved.

Note : We can solve the problems involving the maxima and minima concept by two methods: the first derivative test and the second derivative test. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima