Question

A parent has washed some nappies in a strong bleach solution and wishes to rinse them so that they contain as weak a bleach solution as possible. By wringing out, the nappies can be made to contain just half a litre of solution. Show that two thorough rinses, such that the solution strength is uniform, the first using 12 litres of water and the second using 8 litres of water, reduces the strength of the

1 425
If 20 litres of clean water is all that is available and the parent is prepared to do only two rinses, how best should the water be divided between the two rinses?

Answer 1

Answer: a) Two thorough rinses gives;

1/25 × 1/2 × 2/17 strong bleach/L = 1/425 strong bleach/L

b) The water should be divided into two quantities of 10 liters each

Answer 2

a) Two thorough rinses gives;

1/25 × 1/2 × 2/17 strong bleach/L = 1/425 strong bleach/L

b) The water should be divided into two quantities of 10 liters each

Explanation:

The given parameters are;

The initial volume of strong bleach solution in the nappies = 1/2 Liters

The volume of water first used to rinse = 12 liters

The volume of water used in the second rinse = 8 liters

Therefore, we have;

The total volume of the water and the concentrated bleach in the first rinse = 1/2 + 12 = 12.5 Liters

The new concentration of the bleach in the first rinse water = (1/2 strong bleach)/12.5 L = (1/2 strong bleach)/25/2 L = 1/2×2/25 = 1/25 strong bleach/L

The volume of the first rinse introduced in the second rinse = 1/2 Liters

The concentration of the bleach introduced in the second rinse = The new concentration of the bleach in the first rinse water = 1/25 strong bleach/L

The volume of water added in the second rinse = 8 liters

The total volume of the water and the bleach in the second rinse = (8 + 1/2) liters = 8.5 liters

The concentration of bleach in the second rinse = (The concentration of the bleach introduced in the second rinse × (Volume of bleach solution introduced in the second rinse))/(The total volume of the water and the bleach in the second rinse)

The concentration of bleach in the second rinse = (1/25 strong bleach/L × 1/2 L)/(8.5 Liters)

The concentration of bleach in the second rinse = (1/25 strong bleach/L × 1/2 L)/(17/2 Liters) = 1/25 × 1/2 × 2/17 strong bleach/L = 1/425 strong bleach/L

b) The the quantity of water in the first rinse = x

The amount of water in the second rinse = 20 - x

The concentration of bleach in the first rinse = 1/2/(x + 1/2) = 1/(2·x + 1)

The concentration introduced in the second rinse = 1/2 × 1/(2·x + 1) = 1/(4·x + 2)

The total volume of water and bleach introduced in the second rinse = (20 - x + 1/2) = 20.5 - x

The concentration of bleach in the second rinse = 1/(4·x + 2)/(20.5 - x)

The minimum value for the concentration can be found from taking the derivative of the function for the concentration and equation to zero as follows;

`\frac{\mathrm{d} (1)/(\left ( 4\cdot x + 2 \right )\cdot \left ( 20.5 - x \right ))}{\mathrm{d} x} = (2\cdot \left ( x - 10 \right ))/(\left ( 2\cdot x + 1 \right )^(2)\cdot \left ( 20.5 - x \right )^(2)) = 0`

2·(x - 10) = 0

x = 0/2 + 10 = 10

x = 10

The the quantity of water in the first rinse = x = 10 liters

The the quantity of water in the first rinse = 10 liters

The amount of water in the second rinse = 20 - x = 20 - x = 20 - 10 = 10 liters

The amount of water in the second rinse = 10 liters

The water should be divided into two quantities of 10 liters each

Therefore, the water should be divided into two quantities of 10 liters each to give a final bleach solution concentration of 1/(4·x + 2)/(20.5 - x) = 1/(4×10 + 2)/(20.5 - 10) = 1/42 × 1/10.5 = 1/441 concentration/liter.