Optimizing a Garden Bed
A landscape gardener is designing a rectangular garden bed that will be enclosed by a fence on three sides (the fourth side is against a wall). She has 30 metres of fencing available. She wants to maximize the area of the garden bed. Let the length of the side parallel to the wall be ( x ) metres, and the length of the two sides perpendicular to the wall be ( y ) metres each. Because of space constraints, the side parallel to the wall cannot exceed 20 metres, and each perpendicular side must be at least 2 metres.
1. Write an equation relating \( x \) and \( y \) based on the fencing constraint. Then express the area \( A \) of the garden bed as a function of \( x \) only, and state the domain of \( x \) considering the constraints.
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Fencing used: one side of length ( x ) and two sides of length ( y ), so ( x + 2y = 30 ). Thus ( y = (30 - x)/2 ). Area ( A = x \cdot y = x \cdot (30 - x)/2 = 15x - 0.5x^2 ). Constraints: ( x \le 20 ) (given). Also ( y \ge 2 ) so ( (30 - x)/2 \ge 2 \Rightarrow 30 - x \ge 4 \Rightarrow x \le 26 ), which is automatically satisfied since ( x \le 20 ). Also ( y > 0 ) implies ( x < 30 ), so ( x ) can be from 0 to 20, but since ( y \ge 2 ), ( x ) must be such that ( y \ge 2 ), which gives ( x \le 26 ), so ( x ) in [0,20] is fine. However, ( x ) must be positive to have a garden, so ( 0 < x \le 20 ). Also ( y ) must be positive, so ( x < 30 ), which is true. Thus domain: ( 0 < x \le 20 ).
2. Find the value of \( x \) that maximizes the area, and calculate the maximum area. Verify that this value satisfies the constraints.
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The area function is a downward-opening parabola: ( A(x) = -0.5x^2 + 15x ). The vertex occurs at ( x = -b/(2a) = -15/(2*(-0.5)) = 15 ). So ( x = 15 ) m. Then ( y = (30-15)/2 = 7.5 ) m. Check constraints: ( x = 15 \le 20 ) (ok), ( y = 7.5 \ge 2 ) (ok). Maximum area = ( 15 \cdot 7.5 = 112.5 ) m². Alternatively, compute ( A(15) = -0.5(225) + 15(15) = -112.5 + 225 = 112.5 ). So maximum area is 112.5 m² when ( x = 15 ) m and ( y = 7.5 ) m.