Question

A rectangular field is to be fenced off, and then divided in two by a fence running parallel to one of the sides. If 996 meters of fencing can be used, find the dimensions of the field that will maximize the total area, and then find the maximum area. List the smaller dimension first.

Answer 1

Solution :

Maximize, S = xy

subject to 2x + 2y = 996

So,

2x + 2y = 996

2(x + y) = 996

x + y = 498

y = 498 - x

Therefore,

S = x (498 - x)

S = 498x - $x^2$

S = $-x^2 + 498x$

$S = -(x^2-498x + 249^2) + 300^2$

$S= -(x-249)^2 + 249^2$

S = x (498 - x)

S = 498x - $x^2$

$\frac{dS}{dx}= 498 - 2x$

$\frac{dS}{dx}= 0$

498 - 2x = 0

2x = 498

x = 249

∴ y = 498 - x

  y = 498 - 249

     = 249

Answer 2

Answer:

A = 249 x 249 = 62001 square meter.

Side is L = 249 m

Step-by-step explanation:

Total length of the fence is 996 m.

Let the length is L and the width is W.

Perimeter = 2 (L + W)

996 = 2 (L + W)

498 = L + W

L = 498 - W ...... (1)

Let the area is

A = L W

A = (498 - W) W

A = 498 W - W^2

Differentiate with respect to W.

dA/dW = 498 - 2 W

Put it equal to zero.

498 - 2 W = 0

W = 249 m

Now, L = 498 - 249 = 249 m

Differentiate with respect to W again

$\frac{d^2A}{dW^2} = - 2$

As it is negative so the area is maximum.  

The maximum area is

A = 249 x 249 = 62001 square meter.

Side is L = 249 m