# Section 5.4.1: Single Variable NLP Optimisation Problems

Single variable problems can be dived into those which can conveniently be solved analytically and those which it is best to solve by numerical search.

Problems which appear to be multivariable but include equality constraints may sometimes be reduced to single variable form.

## Example 1

A rectangular platinum catalyst gauze is to provide 300cm¼ of effective area, but requires installation margins of 4cm on each of its longer sides and 6cm on the shorter sides.

Find the dimension which minimises total area.

Let the overall dimensions of the gauze be x cm by y cm. The objective function to be minimised is then clearly:

P = x y

The mounting margins provide the active width of (x-8) and height of (y-12), which must give a total exposed area of 300cm¼ so that there will be an equality constraint:

(x-8) (y-12) = 300

### Solution method

A solution in two variables is in principle possible, but since, as with equation solving, single variable methods are generally much simpler and more robust, we can use each equality constraint to reduce the dimension of the problem by one.

By rearrangement of the equality constraint we can eliminate either of the variable in the o.f. E.g. rearranging to get x:

x = (204+8y) / (4-12y)

The unconstrained single variable o.f. then is:

P = x y = y (204+8y) / (4-12y)

The minimum can be found either by differentiating and setting equal to zero or by a direct search.

## Example 2

The expression for work per unit volume of gas in a 2-stage compressor with intercooling going from pressure P1 to P2 then to P3 is:

W/V = (P1/K) [(P2/P1)K - 2 + (P3/P2)K]

Here K=(k-1)/k where k is the ration of heat capacities, approximately 1.4.

For the case where P1 = 1 atm and P3 = 4 atm confirm the well known result that P2 should be 2 atm.

### Solution method

This is entirely straightforward since clearly W/V should be minimised, and the single decision variable is P2. a numerical search between 1 and 4 atm is most convenient.