A landscape gardener lays a curved border of stones along the graph of y = x²/4 - (ln x)/2, from x…

Step 1. The slope. $f'(x) = \dfrac{x}{2} - \dfrac{1}{2x}$.

Step 2. A perfect square. $1 + [f'(x)]^2 = 1 + \dfrac{x^2}{4} - \dfrac{1}{2} + \dfrac{1}{4x^2} = \left(\dfrac{x}{2} + \dfrac{1}{2x}\right)^2$.

Step 3. The length. $L = \displaystyle\int_1^4 \left(\dfrac{x}{2} + \dfrac{1}{2x}\right)dx = \left[\dfrac{x^2}{4} + \dfrac{\ln x}{2}\right]_1^4 = \left(4 + \ln 2\right) - \dfrac{1}{4} = \dfrac{15}{4} + \ln 2 \approx 4.44$ m.

Step 4. The cost. $4.44\ \text{m} \times 12\ \text{pounds per metre}$ is about $53.32$ pounds. With the unrounded length $4.4431$ m the cost is $53.32$ pounds as well.