5.02b - Inverse proportion

Keeping a product fixed

Can twice as many pumps empty a tank in half the time? Yes, if the tank holds the same amount, every pump works at the same constant rate, and using more pumps does not reduce each pump's output.

Imagine four identical pumps working for three hours. That is twelve pump-hours of work: $4\times3=12$. Six pumps could provide the same twelve pump-hours in two hours. The number of pumps increases, but the time decreases by the reciprocal factor.

The picture represents a fixed total of 24 units. Changing the rectangle's width changes its height, while the product stays 24. This is the same arithmetic as sharing a fixed amount of work.

[DIAGRAM: asset_name: Rectangles with a constant product; asset_slug: inverse-constant-product; caption: Doubling one factor halves the other when their product stays fixed.; description: Two accurately scaled rectangles drawn on the same scale. The left rectangle is four units wide and six units high. The right is eight units wide and three units high. Each has an area label of twenty-four square units, and a central label states width times height equals twenty-four.]

Change only the number of pumps to a larger value. The total work remains fixed.

People are not automatically identical pumps. A claim about workers needs assumptions about equal rates, sharing the task and no delay caused by crowding. A fixed set-up time would also need separate treatment.

Apply the same constant-work reasoning to a fresh task.

The equation of inverse proportion

A fixed product can be written algebraically, so you can find either of the two quantities.

If $xy=k$, divide by nonzero $x$ to obtain $y=k/x$. Conversely, multiplying $y=k/x$ by $x$ gives $xy=k$. These are two forms of the same relationship, with $k$ the constant of proportionality.

At $x=0$, division is undefined. In the pump model, zero pumps do not give a very large finite completion time: the task is never completed by those pumps. Do not insert zero into the formula.