Source: http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx
8. We want to construct a box whose base length is 3 times the base width. The material used to
build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. If the
box must have a volume of 50ft3 determine the dimensions that will minimize the cost to build
the box.
Solution
First, a quick figure (probably not to scale…).
We want to minimize the cost of the materials subject to the constraint that the volume must be
50ft3. Note as well that the cost for each side is just the area of that side times the appropriate
cost.
The two functions we’ll be working with here this time are,
As with the first example, we will solve the constraint for one of the variables and plug this into
the cost. It will definitely be easier to solve the constraint for h so let’s do that.
Plugging this into the cost gives,
Now, let’s get the first and second (we’ll be needing this later…) derivatives,
So, it looks like we’ve got two critical points here. The first is obvious,
, and it’s also
just as obvious that this will not be the correct value. We are building a box here and w is the
box’s width and so since it makes no sense to talk about a box with zero width we will ignore
this critical point. This does not mean however that you should just get into the habit of
ignoring zero when it occurs. There are other types of problems where it will be a valid point
that we will need to consider.
The next critical point will come from determining where the numerator is zero.
So, once we throw out
, we’ve got a single critical point and we now have to verify that
this is in fact the value that will give the absolute minimum cost.
In this case we can’t use Method 1 from above. First, the function is not continuous at one of
the endpoints,
, of our interval of possible values. Secondly, there is no theoretical upper
limit to the width that will give a box with volume of 50 ft3. If w is very large then we would
just need to make h very small.
The second method listed above would work here, but that’s going to involve some calculations,
not difficult calculations, but more work nonetheless.
The third method however, will work quickly and simply here. First, we know that whatever the
value of w that we get it will have to be positive and we can see second derivative above that
provided
we will have
and so in the interval of possible optimal values
the cost function will always be concave up and so
must give the absolute
minimum cost.
All we need to do now is to find the remaining dimensions.
Also, even though it was not asked for, the minimum cost is :
Source: http://tutorial.math.lamar.edu/Classes/CalcI/MoreOptimization.aspx
12. Two poles, one 6 meters tall and one 15 meters tall, are 20 meters apart. A length of wire is
attached to the top of each pole and it is also staked to the ground somewhere between the two
poles. Where should the wire be staked so that the minimum amount of wire is used?
Solution
As always let’s start off with a sketch of this situation.
The total length of the wire is
and we need to determine the value of x that will
minimize this. The constraint in this problem is that the poles must be 20 meters apart and
that x must be in the range
. The first thing that we’ll need to do here is to get the
length of wire in terms of x, which is fairly simple to do using the Pythagorean Theorem.
Not the nicest function we’ve had to work with but there it is. Note however, that it is a
continuous function and we’ve got an interval with finite endpoints and so finding the absolute
minimum won’t require much more work than just getting the critical points of this function.
So, let’s do that. Here’s the derivative.
Setting this equal to zero gives,
It’s probably been quite a while since you’ve been asked to solve something like this. To solve
this we’ll need to square both sides to get rid of the roots, but this will cause problems as well
soon see. Let’s first just square both sides and solve that equation.
Note that if you can’t do that factoring done worry, you can always just use the quadratic
formula and you’ll get the same answers.
Okay two issues that we need to discuss briefly here. The first solution above (note that I didn’t
call it a critical point…) doesn’t make any sense because it is negative and outside of the range
of possible solutions and so we can ignore it.
Secondly, and maybe more importantly, if you were to plug
into the derivative you
would not get zero and so is not even a critical point. How is this possible? It is a solution after
all. We’ll recall that we squared both sides of the equation above and it was mentioned at the
time that this would cause problems. We’ll we’ve hit those problems. In squaring both sides
we’ve inadvertently introduced a new solution to the equation. When you do something like
this you should ALWAYS go back and verify that the solutions that you get are in fact solutions
to the original equation. In this case we were lucky and the “bad” solution also happened to be
outside the interval of solutions we were interested in but that won’t always be the case.
So, if we go back and do a quick verification we can in fact see that the only critical point
is
and this is nicely in our range of acceptable solutions.
Now all that we need to do is plug this critical point and the endpoints of the wire into the length
formula and identify the one that gives the minimum value.
So, we will get the minimum length of wire if we stake it to the ground
pole.
feet from the smaller