Question

A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a​ single-strand electric fence. With 19001900 m of wire at your​ disposal, what is the largest area you can​ enclose, and what are its​ dimensions?

Answer 1

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
`(A)/(x)`.

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

`x+2\left( (A)/(x) \right)`

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

`x+2\left( (A)/(x) \right)=1900 \\ x^2+2A=1900x \\ 2A=1900x-x^2 \\ A=950x- (x^2)/(2)`

For the area to be maximum, the differentiation of A with respect to x must be equal to 0.
i.e.

`(dA)/(dx) =950x- (x^2)/(2)=0 \\ 950-x=0 \\ x=950`

Therefore, the maximum area of the garden enclosed is given by

`A=950x- (x^2)/(2) \\ \\ =950(950)- (950^2)/(2) \\ \\ = (950^2)/(2)= (902,500)/(2) \\ \\ =451,250m^2`

The dimensions of the farmland is 950m by 451,250 / 950 = 950m by 475m