Home > Maths, Starters > Blaise of Glory – a puzzle

Blaise of Glory – a puzzle


This lovely puzzle comes from Chris Smith’s (@aap03102) fantastic maths newsletter. Honestly, if you aren’t on the mailing list get on it.

The puzzle, if you can’t read the picture, is basically this:

There is a concert, if we charge £5 for tickets we sell 120, for every £1 extra we charge we sell ten less tickets, for every £1 less we charge we sell ten more. What should we charge to maximise the profits?

Have you done it? NO?! Well do it NOW, don’t read on till you have. Done? Ok, well this is how I did it:

First I considered the profit itself, given there is no mention of costs, I thought it fair to assume that profit in this case was implied to be the same as turnover (or at least that costs wouldn’t change depending on tickets sold so wouldn’t affect the problem, from here on in I will use the term profit to represent turnover assuming they are equal). So it’s fair to assume that profit is the product of ticket price and tickets sold.

Profit=Ticket Price x Ticket Sales

So I considered ticket prices, it’s £5 +/- an amount. I can model this as

Ticket price = 5 + x,  x is a real number. (x can be positive or negative, and can take any value, although a value less than -5 would mean we are paying people to attend.)

I then considered ticket sales. They are 120 -/+ 10x, (this function will be – 10x when price is + x and vice versa. Which gives:

Ticket Sales = 120 – 10x,  x is a real number less than 12 as you can’t have a negative number if audience members.

This leads to:

Profit = (5 + x)(120 – 10x),  where x is any real number less than twelve.

The final step was to maximise the function. It’s a simple quadratic, my first thought was to expand then complete the square, or use calculus, but then I realised that it would be much easier to use the symmetry of the parabola and just find the midpoint of tge roots. The roots are obvious from our function as it is already in factorised form.

So the roots are x = -5 and x = 12

The midpoint of which is 7/2 or 3.5, which means the maximum occurs when x = 3.5

As x is the number if pounds we have increased the ticket price by, this means to maximise our profits we should charge £8.50

This assumes profit is a continuous function and that a non – integer amount of pounds increase or decrease would increase or decrease the ticket sales proportionally, Ie a 50p increase would decrease sales by 5. This isn’t explicitly stated in the problem, so if only integer increases were allowed then £8 or £9 would give the same profit as each other, which would be tge maximum.

This is an interesting puzzle, and while modelling it I realised that this is, in fact, a great example of a possible use of quadratic equations in the “real world”, outside of the usual areas (ie physics and Engineering.)

  1. February 15, 2015 at 10:20 am

    I just popped it into Excel and out came the answer!

    • February 15, 2015 at 10:30 am


  2. February 15, 2015 at 10:41 pm

    I used:
    sales = 170-10x
    where x = price of a ticket.
    Does that make it a little easier?

    • February 16, 2015 at 7:47 pm

      Aye, thats a nice method.

  1. March 16, 2015 at 5:30 am
  2. March 19, 2015 at 9:03 pm

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