Sustainable Development and Waste Management

A lake's stock of a fish is 5,000 tonnes. Each year, before fishing, the stock grows by 30% through births and growth, minus natural deaths. (Illustrative figures.) (a) What is the largest catch that keeps the stock the same? (b) What happens after two years if 2,000 tonnes are caught every year?

1. (a) Growth $=0.30\times 5000=1500=\; 0.30\; \backslash times\; 5000\; =\; 1500$ tonnes. A catch of 1,500 tonnes a year takes only the growth and leaves the stock at 5,000 t: this is sustainable.
2. (b) Year 1: $5000\times 1.30-2000=45005000\; \backslash times\; 1.30\; -\; 2000\; =\; 4500$ t. Year 2: $4500\times 1.30-2000=38504500\; \backslash times\; 1.30\; -\; 2000\; =\; 3850$ t.

Conclusion: over-fishing shrinks the stock every year, and a smaller stock grows by fewer tonnes, so the decline speeds up. Fisheries use closed seasons, net-size rules and catch limits to keep the catch sustainable.

The amount of organic waste in water can be measured as its biochemical oxygen demand (BOD): the oxygen that bacteria would use to break it down. At a treatment works the BOD was 300 mg per litre in raw sewage, 200 mg per litre after the primary tank and 15 mg per litre in the water released. (Illustrative figures.) (a) What percentage of the BOD was removed in the primary tank? (b) What percentage was removed overall? (c) Which stage removes most, and why?

1. (a) $\frac{300-200}{300}\times 100\mathrm{\%}=33\mathrm{\%}\backslash frac\{300\; -\; 200\}\{300\}\; \backslash times\; 100\backslash \%\; =\; 33\backslash \%$.
2. (b) $\frac{300-15}{300}\times 100\mathrm{\%}=95\mathrm{\%}\backslash frac\{300\; -\; 15\}\{300\}\; \backslash times\; 100\backslash \%\; =\; 95\backslash \%$.
3. (c) The aeration stage: it removes $200-15=185200\; -\; 15\; =\; 185$ mg per litre. Pumping in air lets huge numbers of aerobic bacteria respire fast and digest the dissolved organic matter, which settling alone cannot remove.