# a rectengular field 21 cm^2 in area needs to be fenced. find the dimensions that requires least amount of fence if one side is protected by barn.

*print*Print*list*Cite

To solve your problem, we'll use the fact that for the ratio of the area to the perimeter to be the highest, the sides of the rectangle should be equal. Or in other words it should be a square, as a square is a special case of a rectangle.

Here the area is 21 cm^2. We can have sides equal to sqrt 21 cm and the area would be 21 cm^2. The perimeter in that case would be 4*sqrt 21 cm.

As on one of the sides the area is protected by a barn, we need to apply the fence on the other 3 sides. So we need a length of fence equal to 3*sqrt 21 cm.

**Therefore the least length of fence required is 3*sqrt 21 cm.**

In case it is given that the barn protects the longer side we can use the expression for length of fence as y= x + 2*21/x

dy/dx = 1 - 2*21/x^2

dy/dx = 0 ( for obtaining the minimum value)

=> 1 - 2*21/x^2 = 0

=> x = sqrt [21/2]

**So the length of the fence is 2*sqrt 42**

Whenever we need to determine a maximum or minimum amount of something, we need to create a function that depends on the amount. After creating the function, we'll calculate it's first derivative. Then, we'll calculate the roots of the first derivative. These roots, if they exist, represent the extremes of a function.

In our case, we need to determine the minimum amount to be fenced.

We'll choose as amount one dimension of the rectangle. We'll choose the length and we'll note it as x.

We'll find the other dimension of the rectangle, namely the width, using the formula of area.

A = l*w

We'll substitute area by 21.

21 = x*w

We'll divide by x:

w = 21/x

Now, we'll create the function that depends on x, to determine the minimum amount to be fenced. This function is the perimeter of the rectangle.

The formula of the perimeter of a rectangle is;

P = 2(l+w)

We'll create the function:

P(x) = 2(x + 21/x)

Noe, we'll calculate the first derivative.

P'(X) = (2x + 42/x)'

P'(X) = 2 - 42/x^2

We'll calculate the solution of P'(X).

P'(X) = 0

2 - 42/x^2 = 0

42/x^2 = 2

2x^2 - 42 = 0

We'll divide by 2:

x^2 - 21 = 0

x^2 = 21

x1 = +sqrt21

x2 = -sqrt21

P(sqrt21) = 2sqrt21+ 42/sqrt21

P(sqrt21) = (42sqrt21 + 42sqrt21)/21

**P(sqrt21) = 4sqrt21 cm is the least amount to be fenced. **

Since the rectangular field is protected by a barn on one side we have to take only 3 other sides.

We should consider the least cost of fencing. So for the given the area A = 21 cm^2 (hope it is drawing on a paper) , have to find the perimeter p such that P is least.

3 sides of a rectalngle involves two opposite sides and a side.

Therefore , if x and 21/x are the two adjacent sides of the rectangle with 21 cm^2 area, then the perimeter p = 2x+21/x.

Now to have a minimum value of p , we could go by trial and error method or by calculus method of equating dp/dx to zero and solving for x = x1 and then verify whether the value of {d2p/dx^2 at x=x1} is positive.

dp/dx = (2x+21/x)' = 2-21/x^2

Therefore dp/dx= 0 gives, 2-21/x^2 = 0 . Or 2x^2 = 21. So x1 = sqrt(21/2) = sqrt10.5

d2p/dx^2 = (2-21/x^2)' = 0 -(-2) *21/x^3 = 42/x^3.

So (d2p/dx^2 at x = x1) = 42/ x1^3 = 42/(10.5) is positive. This fulfills the condition that at x= sqrt10.5, the perimeter value is minimum.

Therefore the sides are x = sqrt10.5 , 21/sqrt10.5 and sqrt 10.5 and p = 2* (sqrt10.5)+21/sqrt10.5 = (2*10.5+21)/sqrt10.5 =(42*sqrt10.5)/ 10.5 = 4sqrt10.5 = 12.9615 cm.