A farmer has 120 meters of fencing to enclose a rectangular pasture bordering a straight river (no fence needed along the river). What dimensions maximize the enclosed area?
Answer and explanation
Correct answer: D. Length 60 m parallel to the river and width 30 m perpendicular to the river
Let the width perpendicular to the river be x (two sides) and the length parallel to the river be y (one side). Total fencing: 2x + y = 120 => y = 120 - 2x. Area is A(x) = x y = x(120 - 2x) = 120x - 2x^2. To maximize, take derivative: dA/dx = 120 - 4x = 0 => 4x = 120 => x = 30 m. Then y = 120 - 2(30) = 60 m. Maximum area is 30 60 = 1,800 m^2.
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