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.
Thanks,
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.