Fencing a plot against a wall: completing the square with realistic constraints
Tshepo is fencing a rectangular vegetable plot against the back wall of his yard. The wall forms one whole side of the plot, so he needs fencing for only three sides. He has bought 24 m of fencing and will use all of it. He wants the plot to be at least 5 m wide (the side perpendicular to the wall) so that a wheelbarrow can turn, and he wants the area to be at least 60 m² so that it is worth the effort. He writes the width as w metres and plans to work out the length from the fencing. Fencing costs P85 per metre, and he has already spent P2 040 on it. His neighbour claims that a plot 8 m wide and 8 m long will give the biggest area, since 8 x 8 = 64.
1. Write the area A in terms of w, and use completing the square to find the greatest possible area, stating the width that gives it. Say whether the neighbour's 8 m by 8 m plot is the best choice.
Read the worked answer
Let the side along the wall be L. The fencing covers one length and two widths:
L + 2w = 24, so L = 24 - 2w
A = wL = w(24 - 2w) = 24w - 2w²
Complete the square:
A = -2(w² - 12w) = -2[(w - 6)² - 36] = 72 - 2(w - 6)²
Since 2(w - 6)² is never negative, A can never exceed 72, and A = 72 exactly when w = 6 (then L = 12).
Greatest area = 72 m², with width 6 m and length 12 m.
The neighbour's plot is possible (8 + 8 + 8 = 24 m of fencing) but gives only 64 m². 'Make it square' is the right rule only when all four sides must be fenced; here the free wall changes the best shape.
2. Find the range of widths that satisfy both of Tshepo's conditions (a width of at least 5 m and an area of at least 60 m²).
Read the worked answer
Need 24w - 2w² ≥ 60:
2w² - 24w + 60 ≤ 0
w² - 12w + 30 ≤ 0
Roots: w = [12 ± √(144 - 120)] / 2 = [12 ± √24] / 2 = 6 ± 2.449, i.e. 3.55 and 8.45.
A negative quadratic is below zero between its roots, so 3.55 ≤ w ≤ 8.45.
Combining with w ≥ 5 gives 5 ≤ w ≤ 8.45 (to 2 d.p.), with length 24 - 2w.
Checks: w = 5 gives L = 14 and area 70 m², fine. w = 8 gives L = 8 and area 64 m², fine. w = 8.5 gives L = 7 and area 59.5 m², just too small - so the upper end is real, not an artefact of rounding. Note that the area condition, not the 5 m condition, does most of the work at the bottom end.
3. The neighbour also suggests forgetting the wall and fencing all four sides with the same 24 m 'so the plot is a proper rectangle'. Show what this does to the greatest possible area and explain the result.
Read the worked answer
With four sides fenced, 2(L + w) = 24, so L + w = 12 and L = 12 - w.
A = w(12 - w) = 12w - w² = 36 - (w - 6)²
This is greatest when w = 6, giving L = 6 and A = 36 m².
That is exactly half of 72 m². The explanation: the wall supplies one side free of charge, so the 24 m is spread over three sides instead of four. Spending the same fencing on fewer sides makes every side longer and the enclosed area bigger. The best rectangle is still 'as square as the fencing allows' - two widths of 6 m against one length of 12 m - but the free wall doubles the growing area. The neighbour's suggestion would waste half the potential plot.