Question

A rectangular field is built next to a house and is enclosed by 400 m of fencing for the 3 remaining sides. Determine the dimensions of the field to maximize the area to nearest tenth of a metre.

Answer 1

The dimensions of the field to maximise the area are 100 metres (width) and 200 metres (height).

Explanation:

The formulas for the area (
`A`), measured in square metres, and the perimeter (
`p`), measured in metres, of the rectangle are, respectively:

`A = x\cdot y` (1)

`p = 2\cdot x +y` (2)

Where:

`x` - Width, measured in metres.

`y` - Height, measured in metres.

Note: We assume that height of the rectangle is parallel to the wall of the house.

By (2):

`y = p - 2\cdot x`

In (1):

`A = x\cdot (p-2\cdot x)`

`A = p\cdot x - 2\cdot x^(2)` (3)

Then, we obtain its first and second derivatives by Differential Calculus:

`A' = p - 4\cdot x` (4)

`A'' = -4` (5)

By equalising (4) to zero, we find the following critical value for
`x`:

`x = (p)/(4)`

And besides the Second Derivative Test, this solution is associated to an absolute maximum. Given that
`p = 400\,m`, then the maximum area enclosed by fencing is:

`x = 100\,m`

`A = 20000\,m^(2)`

And the height of the triangle is:

`y = 200\,m`

The dimensions of the field to maximise the area are 100 metres (width) and 200 metres (height).