Question 3e1d2

Dec 14, 2017

Length and width of 900 yards.

Explanation:

Consider that the area of a rectangle can be evaluated as:
Area = Length $\times$ Width

Also consider that the total perimeter of the rectangle must be:
Perimeter = (2$\times$Length) + (2$\times$Width)

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

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

Area $=$ Length$\times$ ($1800 -$Length )
Now we'll let Length = $L$ and Area = $A$ for simplicity:
$\therefore$ A(L) $=$ L$\times$ ($1800 -$L )
$A \left(L\right) = 1800 L - {L}^{2}$

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 $f \left(x\right) = a {x}^{n}$
$f ' \left(x\right) = n a {x}^{n - 1}$

Thus: $A ' \left(L\right) = 1800 - 2 {L}^{1}$
A maximum exists where $A ' \left(L\right) = 0$

$\therefore$ $0 = 1800 - 2 L$
$2 L = 1800$
$L = 900$
Thus for maximum area we must have a length of 900 yards.

Our width would $\therefore$ 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 $\therefore$ be:
Area=$900 \times 900 = 810 , 000$ yards squared.

Dec 14, 2017

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

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

now, $A = l \cdot \frac{3600 - 2 l}{2}$ or ,
$A = \frac{3600 l - 2 {l}^{2}}{2}$
$A = 1800 l - {l}^{2}$

we can take derivative of A wrt l.

$\frac{\mathrm{dA}}{\mathrm{dl}} = 1800 - 2 l$ ---> Eq 2. We will need this again at the end.

When you have reached the maximum or minimum $\frac{\mathrm{dA}}{\mathrm{dl}}$ or any derivative for that matter , it becomes zero.

so we can go ahead and say for the maximum area
$\frac{\mathrm{dA}}{\mathrm{dl}} = 1800 - 2 l = 0$
=> $1800 = 2 l$
=> $900 = l$.
So we got length as 900 , so we can find $b$
$b = \frac{3600 - 2 l}{2} = 1800 - l = 900$
So the length is 900 , breadth is 900.

Now , as said before , when you say $\frac{\mathrm{dA}}{\mathrm{dl}}$ 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 .