Composition of Functions – Applications
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Composition of Functions – Applications

October 11, 2019

So in this video I just
want to talk about a couple of application
problems, or maybe just one application problem
for compositions. So the first problem, or the
problem that I want to look at, in the video is from Calculus
for Business, Economics, and the Social
Sciences by Hoffmann and several other authors. It’s the 11th edition. And the problem goes
something like this. Arthur, the manager of
a furniture factory, finds that the cost of
producing q bookcases– so this is q– during the
morning production run is C of q is equal to q
squared plus q plus $500. So we have a cost function
based on a quantity. On a typical workday,
q of t– that is the quantity as a function
of time– equals 25t bookcases. And they are produced
during the first t hours of a production run
for 0 less than or equal to t is less than or equal to 5. We have a little
statement about the domain of this q of t function. So part a says, “Express the
production cost in terms of t.” So right now I have cost in
terms of quantity, right? I have these two functions. And the reality is,
I want to figure out what the cost is with
relation to time. So this is a composition
of functions, right? q of t gives me an
output that’s quantity, and I can put that output
into this cost function to figure out the
overall function. So I will find C of q of t,
put this 25t in everywhere there’s a q in C.
So let’s do that. So I have 25t squared
plus 25t plus 500. When I square 25t, this
square is on the 25 and the t. So 25 times 25 is 625. Distribute that squared. 25t plus 500. So this is the cost as
a function of time now. So now let’s go to
the next question. How much will have been
spent on production by the end of the third hour? So what’s the cost at the
end of the third hour? I’m going to rewrite this cost
function that we just found, 625t squared plus 25t plus 500. And I am trying
to find cost of 3 when t equals 3, the third hour. So I’ll plug in 3. It’s a time
measurement so I know that it goes into
the time variable. Let me plug that
into my calculator, and you guys do
the same, please. Check me. And I get that the
cost is $6,200. Now, what’s the third question? What’s the average cost of
production during the first 3 hours? And remember, I think we
talked about a problem with average cost is equal
to the cost function, C of whatever, divided
by the quantity. So it’s cost per unit. I’m trying to get an
average cost per each unit. So in this case my
average cost should be C of t divided
by q of t, because I have both of those
in terms of t. And maybe you’re thinking
of this as C of q of t, this composition
function like this. Either way, I want to do
it for the first 3 hours. So I’m really thinking of
C of 3 divided by q of 3. So from the previous problem
I know this one, right? I found the cost at the end of
the third hour to be $6,200. I want to find q of 3. Let’s do that. q of
t is equal to– what was it equal to again? 25t. 25 times 3 is equal to 75. It looks like I can make 75
bookcases in the first 3 hours. So I’ll plug those two numbers
in, 6,200 divided by 75. And I get an average
cost of $82 and it looks like $0.67 per
bookcase if the cost to produce the bookcases
is evenly divided. I think there’s one more
question in this problem. Arthur’s– this
should be budget. There’s a typo there–
allows no more than $11,000 for production during the
morning production run. When will this limit be reached? So I have one quantity
here, $11,000. And I have to figure out
where that number goes in all of the formulas
that I have so far, right? So I have a cost
function, C of q of t. Let me write that down again. That’s 625t squared
plus 25t plus 500. I also have a quantity function,
25t I think it is, right? And so I’m trying to figure
out where does this $11,000 go? q is a quantity of
bookcases that’s in units. t is in time, that
should be in hours. And cost is my only one
that’s in money, right? So I should plug in
this $11,000 right here into this side for
the cost and figure out when that is, right? When will this limit be reached? When is a question that’s
answered by an answer in t, in number of hours, right? That’s a time measurement. When is a time question. So I am going to go ahead
and plug in 11,000 here. And then I get this
quadratic formula or this quadratic equation. So I want to go ahead
and solve here for 0. I’ll subtract 11,000
from both sides. I think I get minus
10,500 on this side. And then 25 is
definitely factorable. You can factor it out
of here, if you want. When I do that, let me divide
10,500 divided by 25 to figure out– this should be
minus 420, this last one. So you can do that. That doesn’t look
like it helps me. I definitely don’t
want to– gosh, I would hate to try
and factor that thing. I’m just going to go ahead
and use the quadratic formula since I know what it is. I know that it’s going to work. And it looks like factoring
is going to be a pain. So let me go ahead and do that. So I have t is
equal to– and this is the negative b plus
or minus the square root b squared minus 4ac over 2a. So t is equal to negative– and
here, since I factor this out, I’m going to go ahead and
use these smaller numbers. Here’s my a is 25, b is positive
1, and c is negative 420. So I’m going to use
those numbers in here. So negative 1 plus or
minus the square root 1 squared minus 4 times 25 times
negative 420 over 2 times 25. Now, you can do these
in different parts, if you want to. I’m going to go ahead
and plug this whole thing into my calculator
to avoid rounding. So I use parentheses. And when I plug in– let’s see,
I’m going to do the plus 1. Negative 1 plus this is
the one that I’m doing. And there’s a 50 down here. If I do this– and I just
want to make this note– if I use this minus
version, negative 1 minus the square
root of whatever, I’ll get a negative
number, right? And you can try that
on your calculator. You get a negative
number, which isn’t really a good output in terms of time. I mean aside from
Back to the Future, we don’t really talk about
negative values of time. So we want to go ahead and
just use the positive value. I guess somehow I thought
the Back to the Future joke was actually funny, when it’s
really not but, you know. So I would go ahead and
put these numbers in. And when I do that, I get
t is equal to 4 point– I think I get a 08 when
I round it– hours. So right around 4 hours
I hit this $11,000. A little bit over 4 hours I
hit the $11,000 mark for cost. So that’s that problem. Let me see. Let’s take a look at
this next problem. It may be a couple questions. This is from a
different calculus book, Essential Calculus by
Wright, Hurd and New and published in 2008. And the problem looks like this. Two students create
a computer program which connects
the dots on a grid so that two players can
play Chase the Rabbit. They buy blank CD discs at a
local store for $0.50 and sell them for $5. They pay another student
$27 per day, 5 days a week, to answer the phone,
take orders, relay customer questions, and make
duplicates of the master disc. And it tells you, let x denote
the number of discs made. Let’s C of x be the weekly
cost function, the total cost, and let R of x be
the revenue function. So the first question says
write an expression for cost, for C of x. So we’ve gone through
an application problem. I encourage you guys, if
you printed this problem out to go ahead and try
it again on your own. Write this cost function. Part b is write the
revenue function. Go through and then come
back and compare and see what you get. So I want to write the
expression for C of x. I’m telling you, you
should push pause now. I know, spoiler alert. I’m about to go to
the problem, right? Write the expression
for C of x, and then what is the weekly cost of
producing 500 discs, so cost. So I have two
parts to the cost I think when I read
through this problem, and sometimes you might have
to read through it twice. But they’re buying
discs for $0.50 each, and they also have a pay out to
this other student of $27 per day. So the cost is the cost of
discs plus the student, right? That’s what they’re
spending money on. So the disc cost $0.50 each
is how I interpret that. So I get 0.50. And let’s say the
number of discs is x. I think that’s
what we said, let x denote the number of discs made. Plus what it costs to pay
the student, $27 a day times 5 days a week. Oops, I put a 5 in there. 27 times 5 days a week. So the cost should be 0.5x
plus– and I think this is 135. Now, the second
part of the question says, what is the weekly
cost of producing 500 discs? And this is what I would call
a C of x function, right? Cost as a function of x. So C of 500, how much does
it cost to produce 500 discs, 0.5 times 500 plus 135. So let me put that
into my calculator. And I get $385. Is that what you guys got? Good. And that seems reasonable to me. You’re making 500 discs. They each cost you $0.50. So that should be around
250 plus this other payout. Part b, write the
revenue function. Revenue– how much money
they’re bringing in, right? So here in this revenue
function we, I think, are bringing in
money from one place. And look back in your
problem, I believe they’re selling the
discs for $5 each. So I did my revenue
function– again, x is the number of discs– is
$5 times however many they sell. So I think that’s
kind of really simple, but I don’t think they’re
making money in any other way. Now, how much
revenue is produced by the sale of 500 discs? R of 500 equals 5 times 500. I think I get $2,500. They make $2,500 by
selling 500 discs. Let’s do part c. Write a profit function. Profit is how much you
bring in– revenue– minus how much you had to
put out– cost, right? So in this case, R
of x minus C of x. Our R of x is 5x and our cost
function is 0.5x plus 135. So let’s simplify this. 5x minus 0.5x minus 135,
that’s 4.5x minus 135, I think, for profit based on
the number of discs. And determine the profit
from the sale of 500 discs. Profit from 500
discs equals– this equals the revenue of 500,
which I’ve already found, minus the cost of 500. Or you could put 500
into the profit function. But I’ve already
plugged these numbers in and I know what they
are, so let me go ahead and put that in here. My revenue is 2,500
and my cost was 385. So let me– on the calculator
I get $2,115 per week, right? This is a weekly profit. So that’s pretty good. And I think that’s the
weekly profit based on sales of 500 discs. D, how many discs must be
sold in order to break even? This idea of breaking even
is kind of what it sounds. It’s the point where you’re
putting out as much money as you’re bringing in. So where R of x is
equal to C of x, this is called your
break even point. So I’m just going to do
that, plug in my functions. Where’s my function? R of x was 5x and the cost
function was 0.5x plus 135. And I’ll just solve for however
many discs I need to sell. So I subtract 0.5
from both sides and then divide by 4.5, 135. And I get x is equal to–
oops, not money, right? x is of quantity is 30 discs. The last one which I would
like you guys to think about, a business professor estimates
that the campus craze– I have a lot of typos
here– craze for the game could become national. And therefore, the game
could be marketed nationally. If 100 colleges were
to become market sites, how might the cost and
revenue functions be affected? Based on your thoughts,
write new functions for cost, revenue, and profit
for the 100 colleges combined together. So I’m not going to do this. Will you guys think about
this and talk about it with your fellow
classmates, how you think that this
would be affected? But that’s the end of
functions and applications of compositions. So let me know if
you have questions.

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