The lengths of sides of the rectangular field should be 40 ft (smaller value) and 60 ft (larger value) respectively.
Let L = length of rectangle area and W = width of rectangle area.
Let the length of side that divides the area be parallel to the width of the area.
The total length of fencing F = 2L + 2W + W
F = 2L + 3W
Since the area is a rectangle, its area is A = LW
Since the area = 2400 square feet,
LW = 2400 ft²
So, L = 2400/W
Substituting L into F, we have
F = 2L + 3W
F = 2(2400/W) + 3W
F = 4800/W + 3W
To find the value at which F is minimum, we differentiate F with respect to W.
So, dF/dW = d(4800/W + 3W)/dW
dF/dW = -4800/W² + 3
Equating dF/dW to zero, we have
-4800/W² + 3 = 0
-4800/W² = - 3
W² = -4800/-3
W² = 1600
W = √1600
W = 40 ft
To determine if this is a value that gives minimum F, we differentiate F twice to get
d²F/dW² = d(-4800/W² + 3)/dW
d²F/dW² = 14400/W³ + 0
d²F/dW² = 14400/W³
substituting W = 40 into the equation, we have
d²F/dW² = 14400/(40)³
d²F/dW² = 14400/64000
d²F/dW² = 0.225
Since d²F/dW² = 0.225 > 0, W = 40 is a minimum point for F
Since L = 2400/W
So, L = 2400/40
L = 60 ft
So, the lengths of sides of the rectangular field should be 40 ft (smaller value) and 60 ft (larger value) respectively.
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