|
Example 1 Find
the dimensions of the box with largest volume if the total surface area is 64
cm2.
Solution
Before we start the process here note that we also saw a
way to solve this kind of problem in Calculus
I, except in those problems we required a condition that related one of
the sides of the box to the other sides so that we could get down to a volume
and surface area function that only involved two variables. We no longer need this condition for these
problems.
Now, let’s get on to solving the problem. We first need to identify the function that
we’re going to optimize as well as the constraint. Let’s set the length of the box to be x, the width of the box to be y and the height of the box to be z.
Let’s also note that because we’re dealing with the dimensions of a
box it is safe to assume that x, y, and z are all positive quantities.
We want to find the largest volume and so the function
that we want to optimize is given by,
Next we know that the surface area of the box must be a
constant 64. So this is the
constraint. The surface area of a box
is simply the sum of the areas of each of the sides so the constraint is
given by,
Note that we divided the constraint by 2 to simplify the
equation a little. Also, we get the
function from this.
Here are the four equations that we need to solve.
(1)
(2)
(3)
(4)
There are many ways to solve this system. We’ll solve it in the following way. Let’s multiply equation (1)
by x, equation (2)
by y and equation (3)
by z. This gives,
(5)
(6)
(7)
Now notice that we can set equations (5)
and (6)
equal. Doing this gives,
This gave two possibilities. The first, is not possible since if this was the case
equation (1)
would reduce to
Since we are talking about the dimensions of a box neither
of these are possible so we can discount . This leaves the second possibility.
Since we know that (again since we are talking about the
dimensions of a box) we can cancel the z
from both sides. This gives,
(8)
Next, let’s set equations (6)
and (7)
equal. Doing this gives,
As already discussed we know that won’t work and so this leaves,
We can also say that since we are dealing with the
dimensions of a box so we must have,
(9)
Plugging equations (8)
and (9)
into equation (4)
we get,
However, we know that y
must be positive since we are talking about the dimensions of a box. Therefore the only solution that makes
physical sense here is
So, it looks like we’ve got a cube here.
We should be a little careful here. Since we’ve only got one solution we might
be tempted to assume that this is dimensions that will give the largest
volume. The method of Lagrange
Multipliers will give a set of points that will either maximize or minimize a
given function subject to the constraint, provided there actually are
minimums or maximums.
The function itself, will clearly have neither minimums or maximums
unless we put some restrictions on the variables. The only real restriction that we’ve got is
that all the variables must be positive.
This, of course, instantly means that the function does have a
minimum, zero.
The function will not have a maximum if all the variables
are allowed to increase without bound.
That however, can’t happen because of the constraint,
Here we’ve got the sum of three positive numbers (because x, y,
and z are positive) and the sum
must equal 32. So, if one of the
variables gets very large, say x,
then because each of the products must be less than 32 both y and z must be very small to make sure the first two terms are less
than 32. So, there is no way for all
the variables to increase without bound and so it should make some sense that
the function, ,
will have a maximum.
This isn’t a rigorous proof that the function will have a
maximum, but it should help to visualize that in fact it should have a
maximum and so we can say that we will get a maximum volume if the dimensions
are : .
|