Raised Garden Bed: Three-Sided Fencing and Soil Volume
The community garden club at a middle school has 40 feet of metal edging to build one rectangular raised bed. The bed will sit against the straight south wall of the building, so edging is needed on only three sides - two widths perpendicular to the wall and one length parallel to it. The club's draft plan is 12 ft wide by 16 ft long, which uses all 40 ft (12 + 16 + 12). A member argues that a square is always the most efficient shape, so the club should build 13⅓ ft by 13⅓ ft instead. The bed must be filled with soil to a depth of 8 inches; soil is sold in bulk by the cubic yard at $34 per cubic yard plus a flat $60 delivery fee, and partial cubic yards cannot be ordered. The principal will fund the project only if the club can show it has the largest possible growing area for the edging it owns.
1. Determine the dimensions that maximize the growing area with 40 ft of three-sided edging. Evaluate the square proposal and the 12 × 16 draft plan against that maximum.
Read the worked answer
Let w be the width perpendicular to the wall. Then the length parallel to the wall is l = 40 − 2w, with 0 < w < 20.
A(w) = w(40 − 2w) = 40w − 2w². This is a downward parabola with vertex at w = −40/(2·(−2)) = 10. So l = 40 − 20 = 20 and A = 10 × 20 = 200 ft². The optimum is a rectangle whose wall-parallel length is twice its width, not a square.
Square proposal (13⅓ × 13⅓): uses 13⅓ + 13⅓ + 13⅓ = 40 ft, area = 177.8 ft² - 22.2 ft² worse. A square maximizes area only when all four sides are fenced with a fixed perimeter; here the wall substitutes for one side, so the balance shifts.
Draft plan (12 × 16): 12 + 16 + 12 = 40 ft, area = 192 ft², which is 8 ft² (4%) below the maximum.
Recommend 10 ft by 20 ft. The member's rule is a special case, not a general principle, and the counterexample of 177.8 ft² versus 200 ft² settles it.
2. Compute the soil volume needed for the optimal 10 × 20 bed at 8 inches deep, in cubic yards, and the total delivered cost. Would building the smaller 12 × 16 draft bed have changed the soil order?
Read the worked answer
Depth in feet: 8 in ÷ 12 = 2/3 ft ≈ 0.667 ft.
- Volume = 200 ft² × 2/3 ft = 133.33 ft³
- Convert: 133.33 ÷ 27 = 4.938 yd³
Since partial cubic yards cannot be ordered, the club must buy 5 cubic yards.
- Soil: 5 × $34 = $170
- Delivery: $60
- Total: $230
Draft bed check: 192 ft² × 2/3 ft = 128 ft³; 128 ÷ 27 = 4.74 yd³, which also rounds up to 5 yd³ - the same $230 order.
That is a useful nuance for the club: the extra 8 ft² of growing area is free in soil cost because both designs fall inside the same purchased cubic yard. Rounding up at the supplier absorbs the difference; if the depth were increased to 10 inches, both would move to 7 yd³ and the comparison would change again.
3. A donor offers 10 more feet of edging, bringing the total to 50 ft. Predict the new maximum area, and explain whether the increase in area is more or less than proportional to the increase in edging.
Read the worked answer
Repeat the model with L = 50: l = 50 − 2w, so A(w) = 50w − 2w², vertex at w = 50/4 = 12.5 ft and l = 25 ft.
A = 12.5 × 25 = 312.5 ft²
Compare: 200 ft² → 312.5 ft² is an increase of 112.5 ft², or 56.25%, from a 25% increase in edging. The gain is more than proportional.
Reason: with three fenced sides the optimum always has w = L/4 and l = L/2, so A = (L/4)(L/2) = L²/8. Area scales with the square of the edging length. Check: (50/40)² = 1.5625, and 200 × 1.5625 = 312.5 ft².
This is worth telling the donor: every added foot of edging is worth progressively more area than the last, so the marginal value of the donation rises. It also means the club should not split the 50 ft between two small beds, which would forfeit the squared scaling.