The Overlapping Gardens
A landscape contractor must fence a rectangular plot of land whose length is 3 m more than twice its width. The plot sits alongside a straight wall, so only three sides need fencing; the wall forms one of the longer sides. Fencing costs $12 per metre, and the contractor has a strict budget of $600. A local rule also requires that the enclosed area be at least 90 m². The contractor writes the width as x metres and the length as 2x + 3 metres, then sets up two inequalities: a cost inequality 12(x + x + 2x + 3) ≤ 600, and an area inequality x(2x + 3) ≥ 90.
1. Solve both inequalities for x and determine the range of widths that satisfies both the budget and the area rule simultaneously.
Read the worked answer
Cost inequality. The three fenced sides are the two widths and one length, so fenced length = x + x + (2x + 3) = 4x + 3. Then 12(4x + 3) ≤ 600 → 4x + 3 ≤ 50 → 4x ≤ 47 → x ≤ 11.75.
Area inequality. x(2x + 3) ≥ 90 → 2x² + 3x − 90 ≥ 0. Factor: (2x + 15)(x − 6) ≥ 0. The roots are x = −7.5 and x = 6. Since width must be positive, the solution is x ≥ 6.
Combining. Both hold when 6 ≤ x ≤ 11.75. Any width in this closed interval meets budget and area rules together; widths below 6 fail the area rule, and widths above 11.75 break the budget.
2. The contractor claims that if the width is exactly 6 m, the budget is comfortably met. Check the claim numerically and explain how the boundary interacts with the area requirement.
Read the worked answer
At x = 6: length = 2(6) + 3 = 15 m. Fenced length = 6 + 6 + 15 = 27 m. Cost = 27 × 12 = $324, which is well under $600 - the claim is correct. Area = 6 × 15 = 90 m², exactly the minimum, so x = 6 is the tight lower boundary of the feasible widths. Note the cost is not the binding constraint at this end; the area rule is. The budget only becomes binding at x = 11.75, where cost = 12(4(11.75) + 3) = 12(50) = $600 exactly.