A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the barn, and the fencing along the west side of the plot is shared with a neighbor who will split the cost of that portion of the fence. If the fencing costs $10 per linear foot to install and the farmer is not willing to spend more than $5000, find the dimensions for the plot that would enclose the most area. (width, length

The dimensions for the plot that would enclose the most area are a length and a width of 125 feet. In this question we shall use the first and second derivative tests to determine the optimal dimensions of a rectangular plot of land. The perimeter ([tex]p[/tex]), in feet, and the area of the rectangular plot ([tex]A[/tex]), in square feet, of land are described below:[tex]p = 2\cdot (w+l)[/tex] (1)[tex]A = w\cdot l[/tex] (2)Where:[tex]w[/tex] - Width, in feet.[tex]l[/tex] - Length, in feet.In addition, the cost of fencing of the rectangular plot ([tex]C[/tex]), in monetary units, is:[tex]C = c\cdot p[/tex] (3)Where [tex]c[/tex] is the fencing unit cost, in monetary units per foot. Now we apply (2) and (3) in (1):[tex]p = 2\cdot \left(\frac{A}{l}+l \right)[/tex][tex]\frac{C}{c} = 2\cdot (\frac{A}{l}+l )[/tex][tex]\frac{C\cdot l}{c} = 2\cdot (A+l^{2})[/tex][tex]\frac{C\cdot l}{c}-2\cdot l^{2} = 2\cdot A[/tex][tex]\frac{C\cdot l}{2\cdot c} - l^{2} = A[/tex] (4)We notice that fencing costs are directly proportional to the area to be fenced. Let suppose that cost is the maximum allowable and we proceed to perform the first and second derivative tests:FDT[tex]\frac{C}{2\cdot c}-2\cdot l = 0[/tex] [tex]l = \frac{C}{4\cdot c}[/tex]SDT [tex]A'' = -2[/tex]Which means that length leads to a maximum area.If we know that [tex]c = 8[/tex] and [tex]C = 4000[/tex], then the dimensions of the rectangular plot of land are, respectively:[tex]l = \frac{4000}{4\cdot (8)}[/tex][tex]l = 125\,ft[/tex][tex]A = \frac{(4000)\cdot (125)}{2\cdot (8)} -125^{2}[/tex][tex]A = 15625\,ft^{2}[/tex][tex]w = \frac{15625\,ft^{2}}{125\,ft}[/tex][tex]w = 125\,ft[/tex]The dimensions for the plot that would enclose the most area are a length and a width of 125 feet. We kindly invite to check this question on areas: