Fencing a Wall-Side Plot
A farmer has 40 m of fencing wire to enclose a rectangular vegetable plot. One side of the plot will use an existing straight wall as a boundary, so wire is needed for only the other three sides. The farmer wants the plot to have an area of 192 m². A neighbour suggests making the side parallel to the wall 24 m long. The farmer instead wants the side parallel to the wall to be at least 20 m. Later, the farmer wonders whether the same 40 m of wire could enclose an area of 210 m² by changing the dimensions. The wire must be used fully, the plot must remain rectangular, and no gate is needed. He is unsure whether both possible shapes are practical and wants the largest possible area checked as well.
1. Find the possible dimensions of the plot that use all 40 m of wire and have area 192 m². Which one meets the farmer's condition that the side parallel to the wall be at least 20 m?
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Let x be the side perpendicular to the wall. Then the side parallel to the wall uses the remaining wire: 40 - 2x. The area is x(40 - 2x) = 192. This gives 40x - 2x² = 192, or x² - 20x + 96 = 0. Factorising: (x - 8)(x - 12) = 0, so x = 8 or x = 12. If x = 8, the parallel side is 40 - 16 = 24 m, giving 8 m × 24 m. If x = 12, the parallel side is 40 - 24 = 16 m, giving 12 m × 16 m. Both use all 40 m and give area 192 m². The condition that the parallel side be at least 20 m selects 8 m × 24 m, so the neighbour's 24 m suggestion is valid.
2. Show whether 210 m² is possible with the same 40 m of wire. Include the largest area that the wire can enclose.
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With the same notation, area A = x(40 - 2x) = -2x² + 40x. Completing the square gives A = -2(x - 10)² + 200. Since -2(x - 10)² is never positive, the maximum area is 200 m², reached when x = 10 and the parallel side is 20 m. Therefore 210 m² is impossible with only 40 m of wire. Algebraically, x(40 - 2x) = 210 gives x² - 20x + 105 = 0, whose discriminant is 400 - 420 = -20 < 0, so there is no real solution. The farmer cannot reach 210 m² without more wire.