We shall look at three problems which involve something be as big as possible (a maximum) or as small as possible (a minimum). For a printable version of the lecture, click here. In the first and third problems, a simple geometrical idea helps to simplify an apparently complicated situation. In the second problem you really need to use calculus to see the full story, but we'll stop short of that here.

The first problem is illustrated here. Try moving the yellow point along the blue line and watch the angle change. Imagine the blue line is a road and the green segment is the front of a building, a stately home perhaps, which you want to photograph from the road. Where is the best place to stand, so that the angle made by the house front at the camera is as large as possible? You can also change the position of the house front if you like, to examine more examples. In the lecture it is explained why the best place to stop is where a circle through the ends of the house front is tangent to (touches) the blue road line. Try this by clicking here.

The key mathematical idea is that, if you move round a circle instead of along a line, then the angle stays the same. You can see this here. You can adjust the size of the circle here too. Notice that a bigger circle gives a smaller, but still constant, angle as you move the yellow point round.

The second problem concerns the distance from a point to a curve, perhaps the distance from a rowing boat to a curved shoreline. Where is the nearest point of the shoreline? In fact click here for an applet which illustrates the graph of the distance function. This shows minima (troughs) and also maxima (peaks) and inflexions, where the graph of distance is momentarily horizontal but continues to flow downwards or upwards none the less. See the text for more information.

The third problem is this: given three towns \(A\), \(B\) and \(C\), where should we place a point \(F\) so that the sum of the distances \(AF + BF + CF\) is as short as possible? In the applet here the triangle \(ABC\) is in green (the letters have been left off to make the picture less cluttered) and the blue triangle is obtained from the green one by rotating about one corner through \(60^\circ\). Try moving the vertices of the green triangle around so that you can verify this.

In this applet you can see the three distance from the point inside the green triangle to the corners; they are coloured yellow, light blue and black. In fact the two black lines are the same length and the two light blue lines likewise. Can you see why this is? To make the required sum as small as possible, we want to move the point inside the green triangle so as to make four points in a straight line. Which four points? Can you see that when this happens, the angles round the point inside the green triangle will be \(120^\circ\)?

Finally a construction for finding the point \(F\) inside \(ABC\) with the angles \(AFB\), \(BFC\), \(CFA\) all \(120^\circ\), is given here. Construct equilateral triangles on the three sides of the given triangle and join corners as shown. It turns out that the three lines meet in a point and this is the required point \(F\).

Of course we can look at more general questions, for example we could start with four towns \(A\), \(B\), \(C\), \(D\) and try to find a point \(F\) inside the region \(ABCD\) so that \(AF + BF + CF + DF\) is as small as possible. See the applet here. In fact if you look at this the right way you can see very easily that the best place for \(F\) is at the meeting point of diagonals \(AC\), \(BD\). Hint: \(AF+FC\) is at least as big as the diagonal \(AC\), and \(BF+FD\) is at least as long as \(BD\).

With four towns we can actually do better, by introducing two additional points, \(F\) and \(G\) say, inside \(ABCD\) and asking for the sum of the lengths of the blue lines here to be as small as possible. This is a hard problem in general, but you might like to consider the special case of \(ABCD\) being a rectangle . Can you find the best positions for \(F\) and \(G\), and show that these achieve a smaller sum of lengths than taking \(F=G\) at the intersection of \(AC\) and \(BD\)? (If you know about calculus you could try using it for this problem.)