Question

Liana has 800 yards of fencing to enclose a rectangular area. How do you maximize the area?

Answer 1

Hint: Now we know that the perimeter of the rectangle is 800 yards. Let us say x is length and y is breadth of rectangle. Hence using this we will find a relation between x and y. Now we know that the Area of rectangle is given by $A=length\times breadth$ . Now we will try to write the equation for Area in y with the help of relation obtained between x and y. Now we know that the condition for maximum is obtained we differentiate the equation and equate it to zero. Hence we will use this to find the condition for maximum.

Complete step by step answer:
Now let us say x is the length of the rectangle and y is the breadth of the rectangle.
Now we are given that the fence is 800 yards long.
Hence the perimeter of the rectangle is 800 yards.
Hence we have
\[\Rightarrow 2\left( x+y \right)=800\]
Now dividing the whole equation by 2 we get,
\[\begin{align}
  & \Rightarrow x+y=400 \\
 & \Rightarrow x=400-y \\
\end{align}\]
Now we want to maximize the area. We know that the area of rectangle is given by $A=length\times breadth$
$\begin{align}
  & \Rightarrow A=x\times y \\
 & \Rightarrow A=\left( 400-y \right)y \\
 & \Rightarrow A=400y-{{y}^{2}} \\
\end{align}$
Now we want to maximize the area hence we will differentiate the equation for Area and equate it to zero to find the condition for extrema.
Hence we get,
$\begin{align}
  & \Rightarrow 400-2y=0 \\
 & \Rightarrow 400=2y \\
 & \Rightarrow y=200 \\
\end{align}$
Now at y = 200 we will get an extrema. Now this extremum cannot be minima as the minimum area will be 0 and the condition for minima will be y = 0. Hence at y = 200 we will get maxima.
Now substituting the value of y in the equation $x+y=400$ we get, x = 200.
Hence the area of rectangle obtained is $200\times 200=40000$

Hence the maximum area is $40000$ sq yards when the length and breadth are equal and is 200 yards.

Note: Now note that for any given perimeter maximum area is given by a square. Also note that differentiating and equating to zero gives us the condition for extrema. Now on differentiating the equation again we get – 2. Now since the second differential is negative we can say that the extremum is maxima by second derivative test.