This next problem is a bit of
a classic and I've actually
gotten multiple requests to do
some variation of this problem.
So it says find the volume of
the largest open box-- It can
be made from a piece of
cardboard-- 24 inches square.
So let me draw my
piece of cardboard.
I don't want to
draw it filled in.
So my piece of cardboard
is going to be a square.
I'll draw it pretty big,
because I want you
to visualize it.
And it's 24, what did they say?
24 inches square.
So it is 24 by 24 square
by cutting equal squares
from the corners and
turning up the sides.
So what are they saying?
Well, they're saying
we're going to cut equal
squares from the corner.
So let's say we cut-- let me
see if I can draw this well
--so I'm going to cut a square
here, and I'm going to cut an
equal size square there-- draw
this neatly --and then cut an
equal size square there,
and then cut an equal
size square there.
So I'm going to cut out those
corners and then I'm going
to fold up the sides.
Just so you can visualize it,
so those dotted lines are where
I would actually fold the box,
and then I would end
up with an open box.
And let's see what the
dimensions would be.
And that's actually what the
whole crux of the problem is.
So how big of a squares should
I cut out in order to maximize
the volume up the box?
Well, these are equal sided
squares that I'm cutting
out, so let's say
that each side is x.
So if this x, then this x, this
is x, this is x, this is x,
this is x, this is x,
this is x, this is x.
And then what is going to be
the volume of our box when
we fold up everything?
Let me see, so I'll
draw the base.
So this dotted square
right here, that's
going to be the base.
I'll draw it at an angle.
That's the base.
And what are the
dimensions of the base?
Well, this base is going to be
the length of this side of the
cardboard minus this x
minus this x, right?
So this base right here, that
is going to be 24 minus 2x.
It's just the whole length
minus this minus that;
that's how long that is.
And of course, this is going to
be the same length, this is
going to be the same length.
All of these sides of the
base are 24 minus 2x.
And then this side is
also 24 minus 2x.
And then what is going to
be the height of this open
box that we're creating?
Let me draw the sides of it.
It's open, which means it
doesn't have a top to it.
So we have an opening here,
so you can kind of see
the inside of the box.
And what is this height?
Well, when you fold these
flaps up the height is x.
If you can think about it--
here this is really just a bit
of visualization --this side
right here, that you can view
as this thing right here.
I can even do it in a
different color for you.
You can view this side,
this front facing side
as this right here.
I'll just do a couple; you
I don't have to do all of
the side, I don't think.
You could view this side
as this side right here.
Right?
It's folded up.
And then this backside is this
backside, and that side there
is that and this is the base.
Anyway, so that is the volume
of our open box and we
want to maximize it.
How do we do that?
Let's write the volume as a
function of x and then take its
derivative, figure out where
the derivative is 0, and hope
that that's a maximum point and
we'll prove it by taking the
second derivative and seeing.
Now we actually want it to be
concave downwards because
concave downwards looks
like that, that means you
found a maximum point.
So we know that the volume as a
function of x-- I like writing
the big v's --the volume of the
function of x is that side
times-- you know, you could
view it as depth times
length times height.
So it's x, the height, times
the depth, 24 minus 2x, times
the width, 24 minus 2x.
Now let's see if I can
multiply this out.
It's probably the hardest
part about this problem.
So x times-- what's
24 times 24?
I want to say it's-- actually
kind of sad, but I've made so
many careless mistakes
recently that I don't
want to make any more.
I want to get the right number.
For 4 times 4 is 16.
4 times 2 is 8, 96.
0.
2 times 24 is 48, 6.
9 plus 8 is 17.
576: that's what I
thought it was.
I should have memorized my
times tables up to 25.
But anyway, 24 times 24 is 576.
And then we'll have 2 times 24,
so that's 48 x, but we have
them twice so it's minus 96x.
Because minus 2x times 24 and
then minus 2x times 24, so
that's 48 to plus
48, so minus 96x.
And then finally the last
two: plus 4x squared.
And now we can multiply the x's
out and we get the volume of
the open box is-- let me put
the x term first --4x to the
third-- I'm just multiplying
this out --minus 96x
squared plus 576x.
Let me erase this.
I should have been able
to do that in my head.
Anyway, I'm not too proud to
show that I didn't know that.
Let's see, so let
me erase that.
And so what do we want to do?
We want to take the derivative
of this, figure out at what x
values do we have a 0 slope,
and then test to see if those
were maximum or minimum points.
If we have a maximum point at
that x value, and if it's a
global maximum, that we have
found the x value that
optimizes the volume.
You might want to graph these
and experiment with them and
get a more intuitive
sense of it.
But that's really what we did
when we found minima and
maxima, and we did concave
upwards and downwards
and all of that.
Anyway, the derivative v prime
of x is equal to 12x squared--
this part is the fun part --12x
squared minus 192x plus 576.
That says 576.
I know it's hard to read;
it's hard to read for me.
So let's figure out
where this equals 0.
So we want to know where
12x squared minus 192x
plus 576 is equal 0.
And the easiest thing to do
here is just divide the whole
thing by 12, both sides.
Because when you divide 0
by 12 you still get 0.
So you get x squared minus--
what's 192 divided by 12?
Well, 192 was 24 times 4.
Is that right?
24?
No, 24 times 16 is 192.
I've been making so many
careless mistakes.
12 goes into 192, 1.
12, 72.
All right.
16.
I should have been
able to do that.
Anyway, minus 16x plus-- well,
this was 24 times 24, so
this should be 12 times 48.
So this should be plus
48 is equal to 0.
And we could write 24
times 24 is 5 plus 76.
So 12 times 48.
Right.
So now we just have
to factor this.
So what two numbers that when I
add them I get minus 16 and
then when I multiply them
I get 48, positive 48?
Let's see 6 times-- I want
to make sure I did this; so
12 goes into 72 six times.
I just want to make sure I
have the numbers right.
As you can tell, I don't do
these problems ahead of time,
because I want you see that I
go through the pain as you do.
So let me make sure.
I have 12 goes into
576 four times.
48, and you get 96.
12 goes into-- right.
48 times, 12 goes into that.
I shouldn't have had to do
that, but I've been making
so many careless mistakes.
I haven't eaten dinner yet.
So anyway, maybe I'm just
missing the easy factor, right?
Because I have, what numbers
you get 12 times-- oh, actually
I just figured it out.
12 times 4.
Sometimes when your brain
is in calculus mode, the
algebra II gets difficult.
So this is the same thing
as x minus 4 times x
minus 12 is equal to 0.
So at x is equal to 4 we're at
some type of minima or maxima
where it's some critical point.
And x is equal to 12.
And let me tell you something:
we don't even have to look at
the secondary derivative here.
What happens when
x is equal to 12?
What will happen to the
volume of the box?
If x is equal to 12, what is
the length of this base?
It's 0.
Because the length of the
base is 24 minus 2x.
So this would be 0,
this would be 0.
So this is the point where we
have a volume of 0, so you
know that this is going
to be a minimum point.
And if you want to verify it,
take the second derivative of
the volume and evaluate the
second derivative of the volume
at 12 and you will realize that
we are concave upwards and that
x equals twelve is
the minimum point.
So you already know, in
all probability, the
answer is x equals 4.
And if you really wanted to
verify it, you can take the
second derivative, add x equals
4, and make sure that we are
concave downwards
at that point.
Because if we want a maximum
point, we want to be some
place where the graph
looks something like that.
So we're at a maximum point.
So what's the
second derivative?
v prime prime of x is
equal to 24x minus 192.
And what's 24 times 4?
It's 96, right?
So v prime prime of
4 is equal to 96.
Right?
4 times 24 minus 192,
which equals minus 96.
So the second derivative at
this point is negative, which
means that we are concave
downwards, which means that
this is a maximum point.
And once again, I challenge you
to find another value of x
where you get a larger value.
And just to get an intuition,
how large will this box be?
Well, if x is 4, then
each side is 24 minus 8.
This would be 16 by 16 by 4.
So that would be the optimal
volume for that box.
Hope you found that useful, and
I will see you in the next
video where I will do another
optimization problem, and
this one is especially fun.
See you soon.
