Question

Find the maximum area of the rectangle that can be formed with fixed perimeter 20 units.

Answer 1

Let the length of the rectangle be l units and breadth be b units. We know the perimeter of the rectangle is 2(l+b) which is given to be 20 units and area is given by $lb$ . So, find the AM and GM of l and b which are $\dfrac{l+b}{2}$ and $\sqrt{lb}$ , respectively. Now we know that AM is always greater than or equal to GM of two positive numbers. Also, for lb to be maximum $\sqrt{lb}$ should be maximum. So, we will take AM=GM and solve to get the maximum value of lb.

Complete step-by-step answer :
We start the solution to the above question by letting the length of the rectangle to be l units and breadth be b units.
Now we know that the perimeter of the rectangle is equal to 2 times the sum of length and breadth.
$\text{Perimeter}=2\left( l+b \right)$
The perimeter is given to be 20 units. So, if we put it here, we get
$20=2\left( l+b \right)$
$\Rightarrow l+b=10......(i)$
We also know that the area of the rectangle is given by the product of length and breadth, i.e., $lb$ .
Now let us find the AM and GM of the two numbers l and b. We know that the AM of two numbers is equal to half of the sum of the two numbers while the GM is root of the product of the two numbers.
So, AM and GM of l and b are $\dfrac{l+b}{2}$ and $\sqrt{lb}$ , respectively.
Now we know that AM is always greater than or equal to GM of two positive numbers and l and b both are positive as they are lengths.
$\therefore \dfrac{l+b}{2}\ge \sqrt{lb}$
Also, for lb to be maximum $\sqrt{lb}$ should be maximum. So, we will take AM=GM and solve to get the maximum value of lb.
$\therefore \dfrac{l+b}{2}=\sqrt{lb}$
Now we will put the value l+b=10. On doing so, we get
$\dfrac{10}{2}=\sqrt{lb}$
$\Rightarrow 5=\sqrt{lb}$
So, we will square both sides of the equation and we know that square of 5 is equal to 25.
$lb=25$
Therefore, the maximum value of lb is equal to 25.
Therefore, we can conclude that the maximum area of the rectangle with perimeter 20 units is 25 sq units.

Note :Remember that for two positive real numbers, $AM\ge GM\ge HM$ . Also, if AM=HM=GM, we can say that both the numbers are equal and the value we are having is the maximum value of their GM and HM, as we did in the above question. If you want you can solve the above equation by the method of application of derivatives as well. Use the equation 2(l+b)=20 to substitute l in terms of b in the equation of area of the rectangle and maximise area by differentiating the equation and putting $dA=0.$