Maximize product within given ranges

[opus]

Homework Statement

A cruise ship that can hold up to 62 people provides excursions to groups of 42 or more.
If the group contains 42 people, each person pays $70.
The cost per person for all members of the party is reduced by $1 for each person in excess of 42.
Find the size of the group that maximizes income for the owners of the ship.

Homework Equations

The Attempt at a Solution

(i) Income = (ticket cost)(number of people)
Let x = the number of people and $p\left(x\right)=70-1x$ is the cost per ticket.
Then,
$I(x)=x⋅p\left(x\right)$

(ii) $I\left(x\right)=x\left(70-1x\right)$
$=-x^2+70x$
$=-\left(x^2-70x+1225\right)+1225$
$=-\left(x-35\right)^2+1225$

(iii) Maximum income is $1225 when the number of people is 35.

Now this answer doesn't make sense, as the minimum amount of people required is 42.
I don't know how to incorporate the range limits of people into the equation. That is, there are only 62 people allowed on the ship, and there is a minimum requirement of 42 people.

Please bare with me. I've been posting a lot of questions, but this is for online classes and there is no lecture to ask questions!

 

[opus]

To summarize my question:
Income = (cost of ticket)(number of people)
But this is for a fixed cost with an unrestricted number of people. I need a fluid cost, and a restricted population.
How can I incorporate these restrictions into my equation for income: $Income=\left(Cost of Ticket\right)\left(Number of People\right)$

 

[Ray Vickson]

Homework Statement

A cruise ship that can hold up to 62 people provides excursions to groups of 42 or more.
If the group contains 42 people, each person pays $70.
The cost per person for all members of the party is reduced by $1 for each person in excess of 42.
Find the size of the group that maximizes income for the owners of the ship.

Homework Equations

The Attempt at a Solution

(i) Income = (ticket cost)(number of people)
Let x = the number of people and $p\left(x\right)=70-1x$ is the cost per ticket.
Then,
$I(x)=x⋅p\left(x\right)$

(ii) $I\left(x\right)=x\left(70-1x\right)$
$=-x^2+70x$
$=-\left(x^2-70x+1225\right)+1225$
$=-\left(x-35\right)^2+1225$

(iii) Maximum income is $1225 when the number of people is 35.

Now this answer doesn't make sense, as the minimum amount of people required is 42.
I don't know how to incorporate the range limits of people into the equation. That is, there are only 62 people allowed on the ship, and there is a minimum requirement of 42 people.

Please bare with me. I've been posting a lot of questions, but this is for online classes and there is no lecture to ask questions!

The ticket cost (per person) is not what you have presented. It should be
$$p(x) = 70 -(x-42),$$
and we must have $42 \leq x \leq 62$.

So, you have in inequality-constrained maximization problem, whose solution will either be obtained by setting a derivative to zero, or else be at one of the two endpoints 42 and 62 (where the derivative need not vanish).

 

[opus]

What was your train of thought to get $p\left(x\right)=70-\left(x-42\right)$? I understand that $70 is the maximum amount a single ticket can cost, so we will be subtracting from that and what we subtract will depend on the number of people x. This x must be $42≤x≤62$ as you stated, but I don't understand the $\left(x-42\right)$

What do you mean by derivative? I googled it and it says it's rate at which the value y changes with respect to x. So like a slope?

 

[opus]

I think I know what you mean with the first part. There is an implied $1 before the parentheses.
$p\left(x\right)=70-1\left(x-42\right)$ and inside the parentheses contains the population that is over the minimum of 42?

 

[Ray Vickson]

What was your train of thought to get $p\left(x\right)=70-\left(x-42\right)$? I understand that $70 is the maximum amount a single ticket can cost, so we will be subtracting from that and what we subtract will depend on the number of people x. This x must be $42≤x≤62$ as you stated, but I don't understand the $\left(x-42\right)$

What do you mean by derivative? I googled it and it says it's rate at which the value y changes with respect to x. So like a slope?

(1) For $x = 42$ the cost is $70 per ticket. For $x = 43$ the cost is $69 per ticket. For $x = 44$ the cost is $68 per ticket, etc. What is the formula you get for
$p(x)$, when p(42) = 70, p(43) = 69, p(44) = 68, etc.?
(2) Never mind about derivatives; just plot the profit as a function of $x$ to see where it is largest in the interval for $x$ from 42 to 62. Your profit curve $P(x)$ vs. $x$ is an inverted parabola whose vertex you can figure out. That should allow you to solve the problem.

 

[LCKurtz]

Please bare with me.

That's "bear". Nudity isn't allowed on this forum.

 

[opus]

That's "bear". Nudity isn't allowed on this forum.

Doesnt mention that anywhere in forum rules.

 

[opus]

(1) For $x = 42$ the cost is $70 per ticket. For $x = 43$ the cost is $69 per ticket. For $x = 44$ the cost is $68 per ticket, etc. What is the formula you get for
$p(x)$, when p(42) = 70, p(43) = 69, p(44) = 68, etc.?
(2) Never mind about derivatives; just plot the profit as a function of $x$ to see where it is largest in the interval for $x$ from 42 to 62. Your profit curve $P(x)$ vs. $x$ is an inverted parabola whose vertex you can figure out. That should allow you to solve the problem.

1) Understood.

2) Once I get an equation for the relationship I have no problems finding the maximum or minimum values. I understand that I am looking for a parabola reflected across the x-axis which will give me some max value p(x) for some number of x people. For this case, the confusion lies in getting the correct equation to graph. To get a maximum or minimum, I know I'll need a quadratic but I'm having trouble actually building the quadratic.
So far we have the "fluid" cost which is $p\left(x\right)=70-\left(x-42\right)$. This is linear, and I need something in terms of the restricted population 42≤x≤62 to get a quadratic. From there, I can find the maximum income.

 

[magoo]

What you have is the cost per ticket: p(x) = 70 - (x-42)
which simplifies as p(x) = 112 - x

Income (i(x)) is simply equal to the price per ticket times the number of tickets or
i(x) = p(x) * x

which is a quadratic

 

[Ray Vickson]

1) Understood.

2) Once I get an equation for the relationship I have no problems finding the maximum or minimum values. I understand that I am looking for a parabola reflected across the x-axis which will give me some max value p(x) for some number of x people. For this case, the confusion lies in getting the correct equation to graph. To get a maximum or minimum, I know I'll need a quadratic but I'm having trouble actually building the quadratic.
So far we have the "fluid" cost which is $p\left(x\right)=70-\left(x-42\right)$. This is linear, and I need something in terms of the restricted population 42≤x≤62 to get a quadratic. From there, I can find the maximum income.

You already have your quadratic, which could be computed for any value of $x$ as a mathematical function. However, it only has practical meaning in this problem for a limited range of $x$. In other words, you just do not look at values of $x$ outside the range, but inside the range you use your quadratic formula. It really is that simple!

Forget about trying to fit the profit formula into constraint $42 \leq x \leq 62$; just restrict the values of $x$ where you use it. That is how these things are done in applications.

 

[opus]

Ok thank you both for the help.