Question

Question #3e1d2

Answer 1

Length and width of 900 yards.

Explanation:

Consider that the area of a rectangle can be evaluated as:
Area = Length `xx` Width

Also consider that the total perimeter of the rectangle must be:
Perimeter = (2`xx`Length) + (2`xx`Width)

But:
We know that the maximum perimeter must be 3600 yards as this is the maximum length of fence that we've got.
`:.` Perimeter`=3600=`(2`xx`Length) + (2`xx`Width)
Rearranging linearly gives:
`=>3600-`(2`xx`Length) =(2`xx`Width)
Dividing both sides by 2 gives:
`:.` `1800-`Length = Width

Consider that this value can now be used within the area equation, allowing for the elimination of one variable.

Area `=` Length`xx` (`1800-`Length )
Now we'll let Length = `L` and Area = `A` for simplicity:
`:.` A(L) `=` L`xx` (`1800-`L )
`A(L) = 1800L - L^2`

This gives us an expression for area in terms of one variable only!

If we now derive this expression with respect to `L` we can evaluate when the function is maximal:

Consider for a polynomial of any power:
If `f(x)=ax^n`
`f'(x)=nax^(n-1)`

Thus: `A'(L) = 1800 - 2L^1`
A maximum exists where `A'(L)=0`

`:.` `0=1800-2L`
`2L=1800`
`L=900`
Thus for maximum area we must have a length of 900 yards.

Our width would `:.` be #`1800-900`=Width from above.
Thus width would also be 900 yards.
NB: Notice how this forms a square?
The maximum area produced would `:.` be:
Area=`900xx900=810,000` yards squared.

Answer 2

length = 900 yards.
breadth = 900 yards.

Explanation:

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 `l` units and a breadth of `b` units.
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`l` + 2`b` = 3600. ---> let this be equation 1.

You have found the relation between `l and b`. Wherever you want , you can express l in terms of b and vice versa.

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
A = `l * b`.
Like stated before , you have to take the derivative of A either with respect to length or breadth . We have both `l and b` here . So what can we do?
we had relation of `l and b` in equation 1.

I have chosen to substitute for `b` as `(3600 - 2l)/2` . (you can always substitute the other way. You just need either this or that in the expression , which makes things easier)

now, `A = l * (3600 - 2l)/2` or ,
`A = (3600 l - 2l^2)/2`
`A = 1800 l - l^2`

we can take derivative of A wrt l.

`(dA)/(dl) = 1800 - 2l` ---> Eq 2. We will need this again at the end.

When you have reached the maximum or minimum `(dA)/(dl)` or any derivative for that matter , it becomes zero.

so we can go ahead and say for the maximum area
`(dA)/(dl) = 1800 - 2l = 0`
=> `1800 = 2l`
=> `900 = l`.
So we got length as 900 , so we can find `b`
`b = (3600 -2l)/2 = 1800 - l = 900`
So the length is 900 , breadth is 900.

Now , as said before , when you say `(dA)/(dl)` is zero , it means it has a chance to be maximum or minimum. So how do you know that you have got the maximum area?

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 .