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

Answer:

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

Step-by-step 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 $x$:

$x = \frac{p}{4}$

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).