Main content

## Solving optimization problems

Current time:0:00Total duration:9:50

# Optimization: box volume (Part 1)

AP Calc: FUN‑4 (EU), FUN‑4.B (LO), FUN‑4.B.1 (EK), FUN‑4.C (LO), FUN‑4.C.1 (EK)

## Video transcript

Let's say that we have
a sheet of cardboard that is 20 inches by 30 inches. Let me draw the cardboard
as neatly as I can. So it might look
something like that. So that is my
sheet of cardboard. And just to make sure
we know the dimensions, there's 20 inches by 30 inches. And what we're
going to do is cut out the corners
of this cardboard. And all the corners are
going to be squares, and we're going to cut
out an x by x corner from each of the corners of this
piece of cardboard-- x by x. Over here, x by x, and
then over here, x by x. And what we'll do is after
we cut out those corners, we can essentially
fold down the flaps. Let me draw the flap. So you could imagine we
can fold right there, we could fold right there,
we could fold right there, and we would form a box. I guess you could imagine a
box without a bottom to it, or you can view a box
without a top to it. So if we were to
fold everything up, we would get a container that
looks something like this. Let me make my best
attempt to draw it. So it would look like this. This is one flap folded up. You can imagine this
flap right over here, if I were to fold
it up like that, it now would look like this. It would now look like this. The height of the flap is x. So this distance
right over here is x. And then if I were to fold
this flap, if I were-- let me do that a little bit neater. If I were to fold this
flap right over here, if I were to fold that up,
then it would look like this. Let me make my best
attempt to draw it. It would look like that. And then I would fold
that back flap up. So the back flap would
look something like that. That would be the back flap. It would look
something like that. And then this flap
over here would-- if I fold it up would
look something like that. And then of course, my
base of my whole thing, so this whole region right over
here of my piece of cardboard, would be the floor of this
box that I'm constructing. And what I want to do
is I want to maximize the volume of this box. I want to maximize
how much it can hold. And I want to maximize it by
picking my x appropriately. So let's think about what
the volume of this box is as a function of x. Well, in order to
do that, we have to figure out all the dimensions
of this box as a function of x. We already know that this
corner right over here, which is made up of when
this side and this side connect when you fold
these two flaps up, that's going to be the
same height over there. That's going to be the
same height over there. The height of this
box is going to be x. But what's the width? What is the width of
this box going to be? Well, the width of
the box is going to be this distance
right over here. And this distance is going
to be 20 inches minus not 1x, but minus 2 x's. So this is going
to be 20 minus 2x. You see it right over here. This whole distance is 20. You subtract this x,
you subtract this x, and you get this
distance right over here. So it's 20 minus 2x. Now, the same logic, what
is the depth of the box? What is that distance
right over there? Well, that distance is this
distance right over here. We know that this entire
distance is 30 inches. If we subtract out this x
and we subtract out this x, we get the distance
that we care about. So this is going
to be 30 minus 2x. So now we have all
of the dimensions. So what would the volume
be as a function of x? Well, the volume
as a function of x is going to be equal
to the height, which is x, times the width, which
is 20 minus x-- sorry, 20 minus 2x times the depth,
which is 30 minus 2x. Now, what are possible values of
x that give us a valid volume? Well, x can't be less than 0. You can't make a
negative cut here. Somehow we would have to add
cardboard or something there. So we know that x is going to
be greater than or equal to 0. So let me write this down. x is going to be greater
than or equal to 0. And what does it
have to be less than? Well, I can cut at
most-- we can see here the length right over here, this
pink color, this mauve color, is 20 minus 2x. So this has got to
be greater than 0. This is always going to be
shorter than the 30 minus 2x, but the 20 minus 2x has to be
greater than or equal to 0. You can't cut more
cardboard than there is. Or you could say that 20 has to
be greater than or equal to 2x, or you could say
that 10 is going to be greater than
or equal to x, which is another way of saying that
x is going to be less than or equal to 10. That's a different
shade of yellow. x is going to be less
than or equal to 10. So x has got to be
between 0 and 10. Otherwise we've cut too
much, or we're somehow adding cardboard or something. So first let's think about
the volume at the endpoints of our-- essentially of
our domain, of what x can be for our volume. Well, our volume when x is
equal to 0 is equal to what? We can have 0 times
all of this stuff. And it's pretty obvious. You're not going to
have any height here. So you're not going
to have any volume, so our volume would be 0. What is our volume
when x is equal to 10? Well, if x equaled 10,
then the width here that I've drawn in
pink would be 0. So once again, we
would have no volume. And this term right over
here, if we just look at it algebraically would
also be, equal to 0, so this whole thing
would be equal to 0. So someplace in between
x equals 0 and x equals 10 we should achieve
our maximum volume. And before we do it analytically
with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. So let me get my TI-85 out. And so first let me set
my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. So my minimum x-value,
let me make that 0. We know that x cannot
be less than 0. My maximum x-value, well,
10 seems pretty good. My minimum y-value,
this is essentially going to be my volume. I'm not going to
have negative volume, so let me set that equal 0. And my maximum
y-value, let's see what would be reasonable here. I'm just going to
pick some a random x and see what type
of a volume I get. So if x were to be 5, it
would be 5 times 20 minus 10, which is 10. So that would be--
did I do that right? Yeah, 20 minus 2
times 5, so that would be 10, and then times
30 minus 2 times 5, which would be 20. So it would be 5
times 10 times 20. So you'd get a volume
of 1,000 cubic inches. And I just randomly
picked the number 5. So let me give my maximum
y-value a little higher than that just in case to
that isn't the maximum value. I just randomly picked that. So let's say yMax is 1,500,
and if for whatever reason our graph doesn't
fit in there, then maybe we can make
our yMax even larger. So I think this is going
to be a decent range. Now let's actually input
the function itself. So our volume is
equal to x times 20 minus 2x minus 2x
times 30 minus 2x. And that looks about right. So now I think we can graph it. So 2nd, and I want to
select that up there, so I'll do Graph. And it looks like we
did get the right range. So this tells us volume is a
function of x between x is 0 and x is 10, and it does look
like we hit a maximum point right around there. So what I'm going
to do is I'm going to use the Trace function
to figure out roughly what that maximum point is. So let me trace this function. So I can still go
higher, higher. OK. So over there my
volume is 1,055.5. Then I can get to 1,056. So let's see, this was
1,056.20, this is 1,056.24, then I go back to 1,055. So at least based
on the level of zoom that I have my
calculator right now, this is a pretty
good approximation for the maximum value that my
function actually takes on. So it looks like my maximum
value is around 1,056, and it happens at
around x equals 3.89. So it looks like
my volume at 3.89 is approximately equal
to 1,056 cubic inches. Or you could say that
we hit a maximum when x is approximately
equal to 3.89. So far, we've just set up
our maximization problem, and we've looked
at it graphically. In the next video, we'll
try to solve it analytically using some of our
calculus tools.

AP® is a registered trademark of the College Board, which has not reviewed this resource.