# Max/Min Problem using derivatives

Question: A professional basketball team plays in an arena that holds 20000 spectators. Average attendance at each game has been 14000. The average ticket price is 75 dollars. Market research shows that, for each $5 reduction in the ticket price, attendance increases by 800. Find the price that will maximize revenue. My work so far (which is not giving me the right answer): Let x be the number of$5 reductions.

R = (75 - x)(14000 + 800x)

R = 1050000 + 46000x - 800x^2

dR/dx = 46000 - 1600x

For max revenue:

when dR/dx = 0:

x = 28.75

when dR/dx DNE:

x = NULL

This is obv. wrong because how can there be 28.75 $5 reductions? Can someone please help out? Thanks! • When you are talking about a continuous/differentiable function, you are talking about a function of a real variable, and it is in this situation that the minimization/maximization problem can be solved in terms of derivatives. What you have is a function of an integer variable and a smooth interpolation of this function – M Turgeon Jan 14 '14 at 5:11 • I don't know what you said but this is what I am supposed to do (Gr 12 Calculus) – Ol' Reliable Jan 14 '14 at 5:13 • In any case, your computation basically shows that the maximum is around the real number 28.75. So which integer is it: 28 or 29? And how can you figure it out? – M Turgeon Jan 14 '14 at 5:15 ## 1 Answer It should be$(75-5x)\times(14000+800x)$. Because decreasing is by \$5 x times.

The number of spectators increases by 800 for every \$5 but not for every \$1.

• ok, but why 5x when i said that x is the number of \$5 reductions? – Ol' Reliable Jan 14 '14 at 5:08
• ok, that makes sense, thanks! – Ol' Reliable Jan 14 '14 at 5:14
• Wait, then how come you don't multiply the 800 by 5? – Ol' Reliable Jan 14 '14 at 5:16
• I got -1.25, why is this negative? – Ol' Reliable Jan 14 '14 at 5:31
• Yes, that is correct. But you know that x is between 0 and 15. Therefore the function is decreasing in the interval [0,15] and you have to take x=0, that is not to change te ticket price in order to have maximum revenue, – kmitov Jan 14 '14 at 5:35