# Extremum problems with rational and root functions · Year 12

> Explanation, worked examples and free practice for the national curriculum in England.

In Extremum problems with rational and root functions (Year 12) you practise 2 skills: Building the function · Problems with a box.

An extremum problem starts with choosing a variable: write the quantity to be made greatest or least in terms of it, and state the domain the variable can move in. In these problems the function can be rational, such as a number plus a multiple of its reciprocal, or have a root, such as the distance between two points.

Then differentiate and set the derivative to 0. Where the derivative is 0 and changes sign there is a turning point: a minimum when it goes from negative to positive, and a maximum when it goes from positive to negative. When a root sits in the denominator of the derivative it is positive, so the numerator decides the sign.

On a closed interval the greatest or least value can also be at an end of the interval, so compare the value at the turning point with the values at the ends. In problems with a box, write the height with the side of the base and the given volume, and the surface area is the sum of the areas of all its faces.

**Tip:** At the end of every problem, go back to the question: sometimes it asks about the variable, and sometimes about the value of the quantity.

## What you learn

### Building the function

- Example: The perimeter of a rectangle is 14 , and one of its sides is x . Write its area as a function of x .
- Answer: −x² + 7x
- Solution: Two neighbouring sides add up to half the perimeter, 7 , so the other side is 7 − x . The area: A(x) = x(7 − x) = −x² + 7x

### Problems with a box

- Example: From a square sheet of side 6 , a square of side x is cut from each corner, and the edges are folded up into an open box. For which value of x is the volume of the box greatest?
- Answer: 1
- Solution: The base is a square of side 6 − 2x , and the height is x . The volume: V(x) = x(6 − 2x)² V(x) = 4x³ − 24x² + 36x The domain: 0 < x < 3 V′(x) = 12x² − 48x + 36 . The derivative is 0 at x = 1 or at x = 3 , and only the first is in the domain. For x < 1 the derivative is positive, and for x > 1 it is negative. This is a maximum point. The volume is greatest when x = 1 .

The exercises are made by rules and the answers are worked out in code. The topic follows selected AS and A level mathematics practice in England (the GCE subject content).

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