0
$\begingroup$

So the question is:

A theatre with $1000$ seats can be filled if tickets are $12$ but for every increase in $25$ cents, the theatre will lose $5$ attendees. What is the cost of the ticket that leads to the maximum profit?

I've found this solution and I'm left with the equation below:

Let $x$ be the number of price increases

$y$ is the total profits

$y = -1.25(x-76)^2 + 19220$

Now my question is, why is $76$ the number of price increases? Is there a special rule that makes $(x-76) 0$?

EDIT:

$y = (12+0.25x)(1000-5x)$

$y = 12000-60x+250x-1.25x^2$

$y = -1.25x^2+190x+12000$

$y = -1.25x(x^2-152x)$

$y = -1.25(x^2-152x+5776-5776)+12000$

$y = -1.25(x^2-152x+5776)+7220+12000$

$y = -1.25(x-76)^2+19220$

$\endgroup$
  • $\begingroup$ @pjs36 I'm not too sure, maybe I should show all of my work. Thanks, though. $\endgroup$ – Erik Jan 7 '17 at 6:06
  • $\begingroup$ Sorry, I misunderstood what you meant by "found the solution" (I thought you meant, this showed up in someone else's solution that you found!) $\endgroup$ – pjs36 Jan 7 '17 at 6:08
0
$\begingroup$

I haven't checked your result, but let's assume you correctly derived the fact that $$y = -1.25(x-76)^2 + 19220.$$

Now suppose $x$ is not $76$. That is, suppose $x - 76 \neq 0$.

If $x - 76 > 0$, then $(x-76)^2 > 0.$ And if $x - 76 < 0$, then $(x-76)^2 > 0.$ That is, whenever $x - 76 \neq 0$, we end up with $(x-76)^2 > 0.$

When $(x-76)^2 > 0,$ then $-1.25(x-76)^2 < 0.$

When $-1.25(x-76)^2 < 0,$ then $-1.25(x-76)^2 + 19220 < 19220.$

But if $x = 76,$ then $x-76=0$ and $-1.25(x-76)^2 + 19220 = 19220.$

In short, if $x = 76,$ then $y = 19220.$ But if $x$ is anything else, then $y < 19220.$

So the highest possible value of $y$ is $19220$ and the only way to get it is if $x = 76$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Putting my equation into a graphing calculator, I finally understood what you meant! Thanks so much! I wish I could accept two answers! Now I can sleep well tonight! Thanks! $\endgroup$ – Erik Jan 7 '17 at 6:33
0
$\begingroup$

$y = (12+0.25x)(1000-5x)$

$\ \ = 12000+250x-60x-1.25x^2$

$\ \ = 12000+190x-1.25x^2$

Since you have to maximise $y$,

$\ \ \ \ \frac{dy}{dx}=0$

$\Rightarrow190-2.5x=0$

$\Rightarrow x=76$

This is the reason why $76$ shows up.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So it has to do with other math concepts? I actually don't know what that means! That's okay though. Thanks! $\endgroup$ – Erik Jan 7 '17 at 6:23
  • $\begingroup$ This concept is for finding maximas and minimas of a given function in calculus. $\endgroup$ – 8hantanu Jan 7 '17 at 6:24
0
$\begingroup$

Consider the standard geometric shifts.

  • replacing $x$ in an expression with $x-a$ creates a horizontal shift of $a$ units,

  • adding $c$ to an expression creates a vertical shift of $c$ units, and

  • multiplying an expression by $k$ scales by $k$

A parabola has exactly one stationary point, and in the most basic form, it is a minimum at the vertex; the negative scaling inverts the parabola and the vertex becomes a maximum.

The basic form can also be described as having a horizontal and vertical shift both of 0, which happens to be the coordinates of the vertex.

If we apply shifts to the parabola, we have in particular shifted the vertex, and since it started at $(0,0)$, it will end up at $(a,c)$.

Thus, the maximum occurs when $x=a$, in your case that is $x=76$ and corresponding value of $y=19220$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! This also makes sense. $\endgroup$ – Erik Jan 7 '17 at 17:13
0
$\begingroup$

Max[y] = max$[-1.25(x-76)^2+19220)]$

= max[$19220 - 1.25(x – 76)^2]$

= max[$19220$ – a never-negative quantity]

= $19220 - 0$ ;if I have a choice of choosing the value of x; and that choice is when x = 76.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.