Length and width of 900 yards.
Consider that the area of a rectangle can be evaluated as:
Area = Length
Also consider that the total perimeter of the rectangle must be:
Perimeter = (2
We know that the maximum perimeter must be 3600 yards as this is the maximum length of fence that we've got.
Rearranging linearly gives:
Dividing both sides by 2 gives:
Consider that this value can now be used within the area equation, allowing for the elimination of one variable.
Now we'll let Length =
This gives us an expression for area in terms of one variable only!
If we now derive this expression with respect to
Consider for a polynomial of any power:
A maximum exists where
Thus for maximum area we must have a length of 900 yards.
Our width would
Thus width would also be 900 yards.
NB: Notice how this forms a square?
The maximum area produced would
length = 900 yards.
breadth = 900 yards.
mona has 3600 yards of fence. She wants the maximum area. So ,she would naturally use all of her fencing.
The fence has to be made a rectangle having a length of
The length of fence is indeed the perimeter of the rectangle she wants.
So perimeter of the rectangle is 2(l+b) units ,and that is equal to 3600 yards.
So , 2
You have found the relation between
Now we want the maximum area possible with perimeter of 3600 yards.
The easier way to maximize/minimize a function(remember ,area is a function of length and breadth) is to express it as derivative with respect to any one of its parameters( like , length or breadth).
area of the rectangle is
Like stated before , you have to take the derivative of A either with respect to length or breadth . We have both
we had relation of
I have chosen to substitute for
we can take derivative of A wrt l.
When you have reached the maximum or minimum
so we can go ahead and say for the maximum area
So we got length as 900 , so we can find
So the length is 900 , breadth is 900.
Now , as said before , when you say
take derivative of Eq 2. wrt to l.
if you get a negative value it means area is maximum . if you get a positive value area is minimum .