I have been set this problem, and although I can derive a minimum perimeter using calculus, I now need to add an extra constraint to one side of the rectangle and I am having problems deriving a suitable equation for the second half of the problem.
An area of 800 $m^2$ is to be fenced off. If the cost per metre of fencing for the frontage is three times as much as the cost per metre of fencing for the remainder, calculate the dimensions for which the cost of fencing is a minimum.
I've worked out that the minimum perimeter is a square 28.28m on a size and the cost of fencing is given by $\textrm{Cost} = 4x + 2y$.
Best Answer
You have posited a rectangle enclosing $800 m^2$. The dimensions are $ x m × {800\over x}m$, where $x$ is deemed to be the length (in meters) of the frontage and the side opposite. Îf it were not for the enhanced cost of the frontal side, then minimizing the perimeter and minimizing the cost would be the same problem -- i.e., find the minimum of the function $2x+{1600\over x}$.
But the fact that the frontal fencing costs treble the fencing for the non-frontal sides changes the function whose minimum is required (so as to minimize the outlay) to $4x+{1600\over x}$ or $4x+1600 x^{-1}$. Differentiating this function gives $4-1600x^{-2}$ or $4-{1600\over x^2}$. Setting this equal to zero and solving for $x$ yields the length of the frontage from which the depth of the rectangle can be found.