The manager of a $1000$ seat concert hall knows from experience that all seats will be occupied if the ticket price is $50$ dollars. A market survey indicates that $10$ additional seats will remain empty for each $ \$5$ increase of the ticket price. What is the ticket price which maximizes the manager's revenue? How many seats will be occupied at that price?

My work, so far:

The possible scenarios are $50\times1000, 55\times990, 60\times980, \ldots$

Let $x =$ ticket price.

Let $y =$ number of tickets sold.

Clearly, $x$ and $y$ are $related$ variables. So, I want to use the method of Lagrange multipliers.

Here are some difficulties, though: $$\text{revenue} = f(x,y) = xy$$

is not a vector valued function $-$ I want to compute the gradient of some vector-valued function, $f$, and solve $$\nabla(f) = \lambda \nabla(g).$$

And moreover, what would the constraint function, $g$, even be?

I can think of two constraints:

$$\max(y) = 1000 \text{ tickets}$$

and $$\min(x) = 50 \text{ dollars}$$

I would like hints only for now.


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    $\begingroup$ This is fundamentally a $1$ variable problem. Increase the ticket price by $x$ increments of $5$ dollars. Then the number of occupied seats is $1000-10x$. Now you can calculate the income $f(x)$ and maximize as usual. $\endgroup$ – André Nicolas Jul 11 '15 at 3:31

As I pointed out in a comment, the problem is quite simply solved using one-variable techniques.

However, let us explore your choice of letting $x$ be the ticket price and $y$ the number of tickets sold. Then the number of $5$ dollar increments is $\frac{x-50}{5}$. The resulting number $y$ of tickets sold is given by $$y=1000-10\cdot \frac{x-50}{5}.$$ This is the constraint, which you may wish to simplify. (There is also the implicit $y\le 1000$.) Now we can use the Lagrange multiplier machinery.

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  • $\begingroup$ Hi @AndreNicolas, first off, thanks so much for the clever suggestion - to turn the problem into a one variable problem, with the variable being the number of price increments instead of the ticket price or tickets sold. I couldn't have thought of that myself and would have still kept trying to set up a two-variable, related variable problem for awhile. I think I have my correct revenue = f(x) set up, I used the usual calculus methods and I'm confident that I have solved it. However, I tried for a long time last night - and now today - to try to use the Lagrange multiplier method, $\endgroup$ – User001 Jul 11 '15 at 22:07
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    $\begingroup$ I set it up for Lagrange multipliers, since that's what you asked for. The constraint simplifies to $2x+y-1100=0$. So we are looking at $xy-\lambda(2x+y-1100)$. Taking partial with respect to $x$, and setting equal to $0$, we get $y-2\lambda=0$. Similarly, $x-\lambda=0$. So $y=2x$. Substitute in $2x+y-1100=0$. We get $x=275$, $y=550$. But as I said, it is easier as a standard first year calculus $1$ variable problem. $\endgroup$ – André Nicolas Jul 11 '15 at 22:17
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    $\begingroup$ In the one variable version where $x$ is the number of $5$ dollar increments, we get after simplification $f(x)=50(1000+90x-x^2)$. Differentiate, set the derivative equal to $0$. We get $x=45$, so the price is $50+5(45)=275$ per ticket. $\endgroup$ – André Nicolas Jul 11 '15 at 22:23
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    $\begingroup$ It is enough to look at the behaviour of the first derivative. The second derivative test is easy here, but often it is not. There are similar techniques available when we are using Lagrange multipliers, but admittedly they are more complicated. Often there is no need for such techniques. From the shape of the problem it may be clear that there is a max, or that there is a min, If we have only one candidate, we know what it is. $\endgroup$ – André Nicolas Jul 11 '15 at 23:00
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    $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Jul 11 '15 at 23:05

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