Let x denote the length of the side of the garden which is bounded by a river, then other (adjacent) side of the farmland has a measure of [tex] \frac{A}{x} [/tex].
The perimeter of a rectangle is given by 2(length + width).
Given
that one of the sides is to be bounded by a river, the the
perimeter of the remaining three sides to be fenced is given by
[tex]x+2\left( \frac{A}{x} \right)[/tex]
Given that there is 1,900 m of wire available to bound the three sides, then the perimeter of the three sides is equal to 1,900
Thus
[tex]x+2\left( \frac{A}{x} \right)=1900 \\ x^2+2A=1900x \\ 2A=1900x-x^2 \\ A=950x- \frac{x^2}{2} [/tex]
For the area to be maximum, the differentiation of A with respect to x must be equal to 0.
i.e.
[tex] \frac{dA}{dx} =950x- \frac{x^2}{2}=0 \\ 950-x=0 \\ x=950[/tex]
Therefore, the maximum area of the garden enclosed is given by
[tex]A=950x- \frac{x^2}{2} \\ \\ =950(950)- \frac{950^2}{2} \\ \\ = \frac{950^2}{2}= \frac{902,500}{2} \\ \\ =451,250m^2[/tex]
The dimensions of the farmland is 950m by 451,250 / 950 = 950m by 475m
