Question

A parking garage owner finds that the demand for 10 hours of parking is modeled by the function, p= -(2/3)x^2 + 9x

where x is the number of parking space in tens of spaces and p is price in dollars. Find the price that should be charged to maximize revenue.

Answer 1

Well, you just have to calculate the extreme values of that function.
You have to calculate p'(x) and then solve for X stating p'(x) = 0

Answer 2

The wording is confusing. Price, p, is a function dependent upon the independent variable, x. So, the question should be, find the number of parking spaces (in units of 10s) that maximizes profit.

One is not free to choose any value for p. The only thing we can change is x, e.g., the objective function to maximize, p(x), presumably factors in such things as the cost of buying space, the cost of securing the space, the cost of cleaning it up, the diminishing returns of having too much space, etc.

Answer 3

It's a calculus question, I was sick the day the teacher taught it so I'm having trouble solving the equation
Approach it as if its a maximize revenue question in calc 1

Answer 4

Yeah, marrionicito already gave you a good response to that. Can you find the derivative?

Answer 5

sure give me a second!

Answer 6

is it -(4/3)x+9?

Answer 7

Yup, so far so good. What's next?

Answer 8

solve for x so set it to 0? 0 = -(4/3)x+ 9 ?

-9 = -(4/3)x => 9/(4/3) = x <=> (27/4) = x ?

Answer 9

This is the correct approach. However, why are we solving p'(x) = 0? I just want to test your understanding.

Answer 10

hmmm well, from what I gather in order to find the maximum revenue we needed to solve for X. In order to have the equation evaluating a change we derived it and set it to zero to find the variable responsible for the increase in marginal revenue ? Should i punch in the derived value for X into the original equation ?

Answer 11

Well, basically, we're solving for p'(x) = 0 because we want to find out where, exactly, p(x) has a slope of 0. Only at this sort of critical point do we have a candidate for a local maximum. Why? Well, because for a function to increase, then decrease, it must go through a point where it is neither increasing nor decreasing.

Now, the second question is, is this point a local maximum or a local minimum? In this case, we know p(x) is concave down (parabola with a negative coefficient in front of the x^2)... in general, we can test to see if the point is a maxima or minima by checking its second derivative at that point. If it's positive, it's a minima, if it's negative, it's a maxima.

A third question is, is this point a global or local maximum? Well, in this case, p(x) only has one critical point, so it must be a local... But if there are multiple, then you must check p(x) at each of those points and choose the largest.

Answer 12

If you see the derivative of a function as a "function that, when you give a value of x*, it gives you back the value of slope of f(x*) then, f'(x) = 0 at some fixed x*. At those fixed x* values, f(x) has a 0 slope. If you "imagine" or "graphic" any continue real to real function, you will see that "the maximums" or "minimums" are the points when the slopes are 0, the function in that point is not growing nor decreasing.

Answer 13

are x values where the slopes of f(x) are 0. and the function in that point is not growing nor decreasing.

Answer 14

Sorry about my english. I´m not get used to write or speak :P

Answer 15

Marrioncito, you're doing quite well. :)

Answer 16

I'm slightly confused Queelius and Marrioncito

Answer 17

so is the answer I got the maximum, or should I insert it into the original equation ? when it comes to the language of the problem I get confused easily. I understand better through examples! :-), you are doing great though! :-)

Answer 18

Verit, you should insert it into the original, yes. Sorry about that. You found the x value that makes p maximum... To find the maximum, do as you indicated, plug that x value into the p function.

Answer 19

okay, thank you! In terms of minimum ... what would be different ?

Answer 20

If it was a minimum, then the second derivative would be positive (the first derivative would still be 0 at the minimum point). In this case, it's a maximum, because the second derivative, p''(x) = -4/3, is negative.

Answer 21

If you're looking for a maximum, choosing the minimum (which may happen by accident if you just solve for p'(x) = 0 and neglect to see if its a min or max) would be the worst possible choice. :)

Answer 22

Ah gotchya! Thank you so much!
In the future, I should do the second derivative test to see if it is a possible minimum, correct?

Answer 23

Yup

Answer 24

What would it mean when p''(x) = 0 for some x? (second derivative is 0)

Answer 25

The concavity is undefined ?

Answer 26

so it wouldn't be known if its a max or min?

Answer 27

Well, that's sort of an interesting idea. But, since you used the word concavity, it already seems like you know what's going on... at p''(x) = 0, it is going from concave up to concave down, or concave down to concave up. Basically, the second derivative is telling us how quickly the slope is changing... if it's 0, then it must be not instantaneously changing at all at that point, i.e., going from increasing slope to no change in slope (that's where p'' is 0) to decreasing slope.

Answer 28

p''(x) = 0 isn't really useful in this sort of min/max problem. I just thought you might benefit from talking more about this stuff. You seem to have a good understanding though, so good job.

Answer 29

thank you ! I usually need dialogue regarding this material because I think too much!
I'm going to close this post, I really appreciate your time and effort Queelius!

Answer 30

bye