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 ([tex]A[/tex]), measured in square metres, and the perimeter ([tex]p[/tex]), measured in metres, of the rectangle are, respectively:
[tex]A = x\cdot y[/tex] (1)
[tex]p = 2\cdot x +y[/tex] (2)
Where:
[tex]x[/tex] - Width, measured in metres.
[tex]y[/tex] - Height, measured in metres.
Note: We assume that height of the rectangle is parallel to the wall of the house.
By (2):
[tex]y = p - 2\cdot x[/tex]
In (1):
[tex]A = x\cdot (p-2\cdot x)[/tex]
[tex]A = p\cdot x - 2\cdot x^{2}[/tex] (3)
Then, we obtain its first and second derivatives by Differential Calculus:
[tex]A' = p - 4\cdot x[/tex] (4)
[tex]A'' = -4[/tex] (5)
By equalising (4) to zero, we find the following critical value for [tex]x[/tex]:
[tex]x = \frac{p}{4}[/tex]
And besides the Second Derivative Test, this solution is associated to an absolute maximum. Given that [tex]p = 400\,m[/tex], then the maximum area enclosed by fencing is:
[tex]x = 100\,m[/tex]
[tex]A = 20000\,m^{2}[/tex]
And the height of the triangle is:
[tex]y = 200\,m[/tex]
The dimensions of the field to maximise the area are 100 metres (width) and 200 metres (height).
