If the rectangular field has notional sides X and y then it has area
A(x) =xy { =6•10^6 sq ft }
The length of fencing required, if
x
is the letter that was arbitrarily assigned to the side to which the dividing fence runs parallel, is:l (x) = 3x +2y
It matters not that the farmer wishes to divide the area into 2 exact smaller areas.
Assuming the cost of the fencing is proportional to the length of fencing required, then
C(x)=a L (x)
To optimise cost, using the Lagrange Multiplier
λ
, with the area constraint :
So the farmer minimises the cost by fencing-off in the ratio 2:3, either-way
