Question

How do you find the maximum area of a rectangle with $80ft$ perimeter?

Answer 1

Hint:
For solving this particular problem we need to differentiate the given function with respect to the independent variable , then equate the required result to zero. By Solving the equation, we get the values for $x$. Substitute the values in the second derivative of the given function. If the result after substitution is less than zero . Then you have to consider it as the maximum point, and if you get the result greater than zero, you have to consider it as minimum.

Complete step by step solution:
According to the question, Perimeter of the rectangle is given as $2(l + b) = 80$, we have
$
   \Rightarrow 2(l + b) = 80 \\
   \Rightarrow 2l + 2b = 80 \\
   \Rightarrow 2l = 80 - 2b \\
   \Rightarrow l = 40 - b..........................(1) \\
 $
Let area of the rectangle be $A = l \times b$
Now substitute $(1)$ in the above equation,
$
   \Rightarrow A = l \times b \\
   \Rightarrow A = (40 - b) \times b \\
 $
          $ = 40b - {b^2}$
We have to find the maximum value; Therefore, we have to differentiate the given function first ,
We will get ,
$A' = 40 - 2b$
Now , set $A' = 0$ ,
We will the following result ,
$ \Rightarrow 40 - 2b = 0$
Subtract $40$from both the side , we will get,
$ \Rightarrow - 2b = - 40$
Multiply by $ - \dfrac{1}{b}$ both the side , we will get ,
$ \Rightarrow b = 20$
Now double differentiate the area function as,
$A'' = - 2$
Here, $A'' < 0$
Therefore, at $b = 20$ we will get the maximum area that is ,
$A = 40b - {b^2}$
$
   = 40(20) - {(20)^2} \\
   = 800 - 400 \\
   = 400 \\
 $

Note:
A function $f(x)$ encompasses a local maximum or relative maximum at $x$ equals to ${x_0}$ if the graph of $f(x)$near ${x_0}$ features a peak at ${x_0}$. A function $f(x)$ features a local minimum or relative minimum at $x$ equals to ${x_0}$ if the graph of $f(x)$ near ${x_0}$ encompasses a trough at ${x_0}$. (To make the excellence clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum.)