The Raised Garden Bed: Area, Edging and a Budget Overrun
A school's agricultural club is building a raised garden bed. The design is a rectangle 6 m long and 4 m wide, with a semicircle joined onto one of the 4 m sides, so that the diameter of the semicircle equals that side. The whole outer boundary of the bed must be edged with timber; there is no wall or fence to use as a free side.
The bed will be filled with imported soil to a depth of 15 cm. Soil is sold only in sealed bags of 0.04 m³ at $4.75 per bag; part bags cannot be bought. Timber edging is sold in straight lengths of 2.4 m at $18.50 each, and pieces are cut from those lengths. The club allows 5% extra timber for cutting waste and joins. Offcuts are discarded.
The club's total budget for soil and timber together is $600 for this bed. The secretary has already ordered 120 bags of soil, saying that 'a little extra soil will not hurt'.
1. Calculate the area of the bed and the total length of timber edging required, correct to 2 decimal places.
Read the worked answer
Area Rectangle = 6 × 4 = 24 m² Semicircle: the diameter is 4 m, so r = 2 m. Area = ½ × π × r² = ½ × π × 4 = 2π ≈ 6.2832 m² Total area = 24 + 6.2832 = 30.28 m²
Perimeter (the edge to be timbered) The boundary is made of: two 6 m sides, one 4 m straight side (the side where the semicircle is not attached), and the curved arc. Arc length = ½ × π × d = ½ × π × 4 = 2π ≈ 6.2832 m Total = 6 + 6 + 4 + 6.2832 = 22.28 m
A common error is to count both 4 m sides of the rectangle. The 4 m side that the semicircle is joined onto is no longer on the boundary - it is replaced by the arc - so it must be left out.
2. How many bags of soil are needed, and what will the soil and timber cost in total? Compare this with the $600 budget.
Read the worked answer
Soil Volume = area × depth = 30.2832 × 0.15 = 4.5425 m³ Bags = 4.5425 ÷ 0.04 = 113.56 → must round up to 114 bags (113 bags give only 4.52 m³, which is short of the 4.5425 m³ required, so 113 is not enough even though it is the 'nearer' whole number.) Cost = 114 × 4.75 = $541.50
Timber Length including waste = 22.2832 × 1.05 = 23.397 m Lengths = 23.397 ÷ 2.4 = 9.75 → round up to 10 lengths Cost = 10 × 18.50 = $185.00 (10 lengths give 24 m, so 0.60 m is spare - within the 5% allowance, so no extra length is needed.)
Total = 541.50 + 185.00 = $726.50
The project is over budget by 726.50 − 600 = $126.50. Neither rounding choice caused this: the overrun comes from the size of the bed itself.
3. Evaluate the secretary's decision to order 120 bags, and recommend a change that brings the project within the $600 budget.
Read the worked answer
The 120-bag order Cost = 120 × 4.75 = $570.00. With timber at $185.00 the total is $755.00, which is $155 over budget.
It is also wasteful on its own terms. The job needs 114 bags, so 6 bags - 0.24 m³ of soil worth 6 × 4.75 = $28.50 - are surplus. Buying 'a little extra' pushed the project $28.50 further into deficit without improving the bed.
Finding a solution within budget Timber is fixed at $185, so soil must cost no more than 600 − 185 = $415. That allows 415 ÷ 4.75 = 87.4 → 87 bags = 87 × 0.04 = 3.48 m³. Maximum depth = 3.48 ÷ 30.2832 = 0.115 m, so 15 cm is impossible.
A practical fix is a 10 cm depth: Volume = 30.2832 × 0.10 = 3.0283 m³ → 3.0283 ÷ 0.04 = 75.7 → 76 bags Cost = 76 × 4.75 = $361.00 Total = 361.00 + 185.00 = $546.00, leaving $54 of the budget spare.
Recommendation: cancel the 120-bag order, buy 76 bags and build the bed to 10 cm. Note the trade-off: a shallower bed dries out faster and suits shallow-rooted crops only, so the club should confirm that 10 cm is deep enough for what it intends to plant before changing the order.