The dimensions that will maximize the area are x = 354 ft and y = 708 ft
The length of the side rectangle perpendicular to the river is 354 ft
The length of the side of the rectangle parallel to the river is 708 ft
The maximum area = 250632 ft²
Explanation:Given:
The length of the fencing = 1416 ft
The length of the side rectangle perpendicular to the river = x
To find:
The dimensions that will maximize the area
To determine the dimensions, we will make an illustration of the given information:
let the length o the rectangle parallel to the river = y
Length of the for the enclosed area = Perimeter of the enclosed area
Perimeter of the enclosed area = x + x = y = 2x + y
[tex]1416=2x+y\text{ . . .\lparen1\rparen}[/tex]Area of the rectangle = length × width
length = y, width = x
let the Area of the rectangle = A(x)
[tex]A(x)\text{ = xy . . . \lparen2\rparen}[/tex]To get the expression for A(x), we will make y the subject of the formula in equation (1):
y = 1416 - 2x
substitute for y in equation (2):
[tex]\begin{gathered} A(x)\text{ = x\lparen1416 - 2x\rparen} \\ \\ A(x)\text{ = 14166x - 2x}^2 \end{gathered}[/tex]To get the maximum dimension, we will differentiate with respect to x:
[tex]\begin{gathered} A^{\prime}(x)\text{ = 1416 - 4x} \\ \\ At\text{ maximum, A'\lparen x\rparen = 0:} \\ 1416\text{ - 4x = 0} \\ 1416\text{ = 4x} \\ x\text{ = }\frac{1416}{4} \\ x\text{ = 354} \end{gathered}[/tex]substitute for x in equation (1):
[tex]\begin{gathered} 1416\text{ = 2\lparen354\rparen + y} \\ 1416\text{ - 708 = y} \\ y\text{ = 708} \end{gathered}[/tex]The dimensions that will maximize the area are x = 354 ft and y = 708 ft
The length of the side rectangle perpendicular to the river is 354 ft
The length of the side of the rectangle parallel to the river is 708 ft
The maximum area = 354 × 708
The maximum area = 250632 ft²
