A farmer has 400 fet of fening wants to endlosc a rectangula

A farmer has 400 fet of fening wants to endlosc a rectangular pen and then A farmer has 4000 feet of fencing wants to enclose a rectangular pen and then divide it into two plots with a fence parallel to one of the pairs of sides (see figure below). He does not need to fence the side next to the barn. What are the dimensions of the pen that will yield a maximum area? What is the maximum area that can be enclosed? (3 poinks each) No Fence

Solution

Let the length and the width of the plot be x and y ft. respectively. Then the area of the plot is A = xy. Also x + 3y = 4000. Then A = x(4000-x)/3 = 1/3(4000x - x²) . We know that A is maximum when dA/dx = 0 and d²A/dx² is negative. Here, dA/dx = 1/3(4000- 2x) which is 0 when 2x = 4000 , i.e. when x = 2000. Also, d²A/dx² = -2/3 so that for maximum area, we must have x = 2000. Then, 3y = 4000 -x = 4000 -2000 = 2000 so that y = 2000/3. Then the maximum area is A = xy = 2000*2000/3 = 4000000/3 = 1333333.33 sq. ft.