a rectangular are with two additional fences across its widt

a rectangular are with two additional fences across its width must be constructed. find the maximum area that can be enclosed with 120 m of fencing. ansswer= 450

thanks! please show work, answer is given to check work

Solution

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Let the length and the breadth of the rectangular area be x and y meters respectively and llet A be the area. Then, A = xy...(1). Also, the perimeter of the rectangular area is 2x + 2y. Since two addituional fences are required across the width, the total requirement of fencing is 2x + 2y + 2y = 2x + 4y  Since the total fencing is 120 meters, therefore 2x + 4y = 120 or, x + 2y = 60 or, y = 1/2 ( 60 - x)...(2) . On substituting y = 1/2 ( 60 - x) in the 1st equation, we have A = x ( 60 -x) /2 = 30x - x²/ 2. Then dA/ dx = 30 - x and d²A/ dx² = 1. Now, A is maximum when dA/dx = 0 and d²A/dx² is negative. Since  d²A/dx² = -1, A will be maximum when 30 - x = 0 i.e. when x = 30. Then from the 2nd equation , we have y = 1/2 ( 60 -x) = 1/2 ( 60 - 30) = 30/2 = 15. Thus the maximum area A = 30 * 15 = 450 square meters.