Video Transcript
In this video, we will see one of
the practical applications of differentiation to optimization problems. That is, problems where we’re asked
to determine the maximum or minimum value of a given function such as time, area, or
perimeter. We’ll be able to answer questions
like what does the maximum area I can enclose with a given length of fencing or what
is the highest point reached by a rocket following a given curve. In some examples, we’ll also see
how to form the optimization function and any constraints ourselves from a worded
description.
We recall, first of all, some key
facts about using differentiation to find the critical points of a function. The critical points of a function
are those points in its domain where its first derivative, 𝑓 prime of 𝑥, is equal
to zero or is undefined. To find the 𝑥-value at a critical
point, we can differentiate the function to find its gradient, or slope, function
and then set this equal to zero and solve the resulting equation. We can find the value of the
function itself at the critical point by substituting our 𝑥-value or values into
the function.
We must also remember to confirm
that our critical point is indeed a maximum or a minimum by applying the second
derivative test. We recall that at a local maximum,
the second derivative 𝑓 double prime of 𝑥 will be negative, whereas at a local
minimum, the second derivative 𝑓 double prime of 𝑥 will be positive. We’ll now see how to apply these
key principles to some optimization problems.
A rocket is launched in the
air. Its height, in meters, as a
function of time is given by ℎ of 𝑡 equals negative 4.9𝑡 squared plus 229𝑡 plus
234. Find the maximum height the rocket
attains.
So, we’ve been given an equation
for the height ℎ of this rocket in terms of 𝑡 time. And we’re asked to determine the
maximum height that the rocket attains. We can follow some key steps in
order to do so. First, we need to find an
expression for the first derivative of our function. In this case, that’s ℎ prime of
𝑡. By applying the power rule of
differentiation, we see that ℎ prime of 𝑡 is equal to negative two multiplied by
4.9𝑡 plus 229, which simplifies to negative 9.8𝑡 plus 229.
Next, we recall that at the
critical points of the function, the first derivative is equal to zero. So, we’re going to take that
expression we’ve found for ℎ prime of 𝑡, set it equal to zero, and then solve the
resulting equation for 𝑡. We add 9.8𝑡 to both sides and then
divide by 9.8 to give 𝑡 equals 229 over 9.8, which, as a decimal, is 23.36734, or
23.37 to two decimal places. Now, this is the value of 𝑡 at
which our function ℎ of 𝑡 has a critical point. We don’t yet know whether it is a
maximum. And we don’t yet know the height
that the rocket attains at this point.
Our next step, then, is to evaluate
the function ℎ of 𝑡 when 𝑡 is equal to 23.37. Substituting into our equation ℎ of
𝑡 gives ℎ of 23.37 equals negative 4.9 multiplied by 23.37 squared plus 229
multiplied by 23.37 plus 234. Which evaluates to 2909.56119, or
2909.56 correct to two decimal places. So, we believe that this is the
maximum height that the rocket attains as this occurs at the only critical point of
the function. But we must confirm that it is
indeed a maximum.
To do so, we’ll perform the second
derivative test. We’ll evaluate the second
derivative ℎ double prime of 𝑡 at this critical point. Differentiating our expression for
ℎ prime of 𝑡, which was negative 9.8𝑡 plus 229, we find that ℎ double prime of 𝑡
is equal to negative 9.8. And in fact, we don’t need to
evaluate the second derivative when 𝑡 equals 23.37 because it is a constant. The second derivative is the same
for all values of 𝑡.
We note though, that this value is
negative. And therefore, by the second
derivative test, our critical point is a maximum. Now, we could also have seen this
if we considered that the expression we were given for ℎ in terms of 𝑡 is a
quadratic expression with a negative leading coefficient. And therefore, the graph of 𝑡
against ℎ would be an inverted parabola. And we know that it will therefore
have a maximum point rather than a minimum. So, we can conclude that the
maximum height the rocket attains, and we have confirmed that it is indeed a
maximum, is 2909.56 meters, correct to two decimal places.
Now, it’s just worth pointing out
that in this question, each of the derivatives also have practical
interpretations. The first derivative of our
function, that’s negative 9.8𝑡 plus 229, gives the velocity of the rocket, which we
see is decreasing as time increases. The second derivative, which we saw
is just the constant negative 9.8, gives the acceleration of the rocket, or in fact
the deceleration. Which, we see, is equal to negative
9.8 meters per second squared. And that’s the deceleration due to
gravity.
In this problem, the function we
needed to optimize was given to us. But it will more often be the case
in optimization problems that we’ll need to form these functions ourselves from a
worded description. Let’s see how this works in our
next example.
Find two numbers whose sum is 156
and the sum of whose squares is the least possible.
Let’s allow these two numbers,
which we don’t yet know, to be 𝑥 and 𝑦. Then, we can express the fact that
their sum is 156 as 𝑥 plus 𝑦 equals 156. We want to minimize the sum of
their squares, which we can call 𝑠. 𝑠 is equal to 𝑥 squared plus 𝑦
squared. In order to do so, we need to find
the values of 𝑥 and 𝑦 for which the rate of change of 𝑠 with respect to either 𝑥
or 𝑦, is equal to zero. This means that we need to
differentiate 𝑠 with respect to either 𝑥 or 𝑦.
First, though, we need to write 𝑠
in terms of one variable only. The choice is entirely arbitrary in
this problem. We could perform a simple
rearrangement of our first equation to give 𝑦 equals 156 minus 𝑥 and then
substitute this expression for 𝑦 into our equation for 𝑠 to give an equation in
terms of 𝑥 only. Distributing the parentheses and
then simplifying gives 𝑠 is equal to two 𝑥 squared minus 312𝑥 plus 24336.
Remember, we’re looking to minimize
this sum of squares, so we need to find the critical points of 𝑠. To do so, we need to find where the
first derivative of 𝑠 with respect to 𝑥, that’s d𝑠 by d𝑥, is equal to zero. We can use the power rule to find
this derivative. And we see that d𝑠 by d𝑥 is equal
to four 𝑥 minus 312. We then set this derivative equal
to zero and solve for 𝑥. We first add 312 to each side and
then divide by four, giving 𝑥 equals 78.
So, we found the value of 𝑥 at
which 𝑠 has a critical point. But there are two things we need to
do. In a moment, we’ll confirm that
this is indeed a minimum. But first, we also need to find the
value of 𝑦, which we can do by substituting the value of 𝑥 into our linear
equation. We see that 𝑦 is equal to 156
minus 78, which is equal to 78.
To confirm that this critical point
is indeed a minimum. We need to find the second
derivative of the function 𝑠 with respect to 𝑥. Differentiating d𝑠 by d𝑥 again
gives d two 𝑠 by d𝑥 squared is equal to four. The second derivative of 𝑠 with
respect to 𝑥 is, therefore, constant for all values of 𝑥. And more importantly, it is
positive, which confirms that this critical point is indeed a minimum. The two numbers then whose sum is
156, which have the minimum sum of squares, are 78 and 78.
In this example, we saw how to set
up the optimization function and any constraints ourselves from information given in
the question. We’ll, now see how to do this again
with a more practical example.
A wire of length 41 centimeters is
used to make a rectangle. What dimensions give its maximum
area?
Now, it’s important that we don’t
just attempt trial and error here. We need to use a proper
optimization method to find the dimensions which will give the largest possible area
for this rectangle, subject to the constraint that we only have 41 centimeters of
wire.
Let’s consider then a rectangle
with a length of 𝑙 centimeters and a width of 𝑤 centimeters. We need to maximize its area, which
for a rectangle is its length multiplied by its width. Subject to the constraint which is
that the perimeter of this rectangle must be equal to 41. The perimeter of a rectangle is
twice its length plus twice its width. So, we have the constraint two 𝑙
plus two 𝑤 equals 41.
Now, in order to maximize this
area, we’re going to need to use differentiation to find the critical points of this
function 𝐴. But before we can do that, we need
to write 𝐴 in terms of one variable only. The choice of whether we use 𝑙 or
𝑤 is entirely arbitrary. So, I’ve chosen to rearrange the
linear equation to give 𝑙 equals 41 minus two 𝑤 over two. Substituting this expression for 𝑙
into our area formula gives 𝐴 equals 41 minus two 𝑤 over two multiplied by 𝑤. And distributing the parentheses,
we have 41𝑤 over two minus 𝑤 squared.
To find the critical points of 𝐴,
we first find its first derivative, d𝐴 by d𝑤, which, using the power rule of
differentiation, is equal to 41 over two minus two 𝑤. We then set this expression equal
to zero and solve the resulting equation for 𝑤. We add two 𝑤 to each side and then
divide by two, giving 𝑤 is equal to 41 over four. So, we’ve found the width of the
rectangle at which the area has a critical point.
We also need to find the length
though, which we do by substituting this value of 𝑤 back into our expression for
𝑙, giving 41 minus two times 41 over four all over two. Which also simplifies to 41 over
four. However, we’re not yet
finished. We know that these values of 𝑤 and
𝑙 give a critical point for the area. But we haven’t yet confirmed that
it is indeed a maximum. To check this, we need to perform
the second derivative test. We find d two 𝐴 by d𝑤 squared,
which is equal to negative two.
Now, this is a constant for all
values of 𝑤. But more specifically it is a
negative constant. And as the second derivative is
less than zero, this confirms that our critical point is indeed a maximum. So, we found that the length and
width which give this rectangle its maximum area, subject to the given perimeter
constraint, as decimals, are both 10.25 centimeters.
Now, we notice that the length and
width of this rectangle are actually the same, making it in fact a square. This illustrates a general point in
optimization problems where we’re looking to maximize an area with respect to a
length constraint. The maximum area will be achieved
when the dimensions are as similar as possible i.e. when the ratio between the
dimensions is as close as possible to one to one.
In the case of a problem involving
a rectangle, it will always turn out that the shape will in fact be a square. But of course, we must always go
through the working out. In order to show this. In our final example, we’ll see how
to maximize the sum of the volumes of two three-dimensional shapes subject to a
surface area constraint.
Given that the sum of the surface
areas of a sphere and a right circular cylinder is 1000𝜋 centimeters squared, and
their radii are equal, find the radius of the sphere that makes the sum of their
volume at its maximum value.
So, in this question, we’re asked
to maximize the sum of the volumes of two three-dimensional solids, subject to a
constraint about the sum of their surface areas. Let’s begin by writing down fomulae
for the surface areas of both the sphere and the right circular cylinder. And as their radii are the same, we
can use the same letter 𝑟 for both. For the sphere, first of all, its
surface area is given by four 𝜋𝑟 squared. For the cylinder, its surface area
is two 𝜋𝑟 squared plus two 𝜋𝑟ℎ, where ℎ represents the height of the
cylinder.
As the sum of these surface areas
is 1000𝜋 centimeters squared, we can form an equation, four 𝜋𝑟 squared plus two
𝜋𝑟 squared plus two 𝜋𝑟ℎ equals 1000𝜋. We can then combine the like terms
on the left-hand side and then divide through by a 𝜋, as this is a common factor in
all terms. We could also divide through by
two, as all the coefficients are even, to give three 𝑟 squared plus 𝑟ℎ equals
500. We can’t do anything further with
this equation at this point, as we have two unknowns 𝑟 and ℎ. So, next, we recall the formulae
for the volume of a sphere and the volume of a cylinder.
The volume of a sphere is
four-thirds multiplied by 𝜋 multiplied by its radius cubed. And the volume of a cylinder is 𝜋
multiplied by its radius squared multiplied by its height. So, we have that the total volume
of these two solids is four-thirds 𝜋𝑟 cubed plus 𝜋𝑟 squared ℎ. We want to maximize the sum of
these volumes, 𝑉 total. Now, this will be maximized when
its rate of change with respect to either 𝑟 or ℎ is equal to zero, which will be
when its first derivative is equal to zero. But before we can differentiate, we
need to express 𝑉 total in terms of a single variable.
It’s much more straightforward to
rearrange our surface area constraint to give an expression for ℎ in terms of 𝑟
than it is to give an expression for 𝑟 in terms of ℎ. We have ℎ equals 500 minus three 𝑟
squared over 𝑟. We can then substitute this
expression for ℎ into our expression for the total volume so that it is in terms of
𝑟 only. We can cancel a factor of 𝑟 in the
second term and then distribute the parentheses to give four-thirds 𝜋𝑟 cubed plus
500𝜋𝑟 minus three 𝜋𝑟 cubed. We have an expression for 𝑉 total
in terms of 𝑟 only.
Next, we need to find the first
derivative d𝑉 total by d𝑟, so we’ll create a little bit of space in order to do
this. By applying the power rule of
differentiation, we see that the derivative of 𝑉 total with respect to 𝑟 is equal
to four-thirds 𝜋 multiplied by three 𝑟 squared plus 500𝜋 minus three 𝜋
multiplied by three 𝑟 squared, which all simplifies to 500𝜋 minus five 𝜋𝑟
squared. Next, in order to find critical
points, we need to set this derivative equal to zero and solve for 𝑟.
We can divide through by five 𝜋,
giving zero equals 100 minus 𝑟 squared. Adding 𝑟 squared to both sides
gives 𝑟 squared equals 100. And we then find 𝑟 by square
rooting. We only need to take the positive
square root as the radius of a solid must be a positive value. So, we see that 𝑟 is equal to
10. We now know that the combined
volume of these two solids has a critical point when the radius is equal to 10. But we must now confirm that it is
a maximum.
We perform the second derivative
test. Differentiating our expression for
d𝑉 total by d𝑟 again, with respect to 𝑟, gives negative 10𝜋𝑟. And evaluating this when 𝑟 is
equal to 10 gives negative 100𝜋. This is negative, which confirms
that the critical point is indeed a maximum. So, we found that the radius of the
sphere and also the radius of the right circular cylinder which maximizes the sum of
their volumes, subject to the given surface area constraint, is 10 centimeters.
Let’s summarize then the key points
that we’ve seen in this video. Firstly, the key principles of
differentiation can be applied to optimization problems. That’s problems where we want to
find the maximum or minimum value of a function. We know that the critical points of
a function occur when its first derivative is equal to zero or is undefined. Once we found a critical point, we
must always confirm that it is indeed a maximum or minimum by performing the second
derivative test. And we’ve seen that it may be
necessary for us to form both the optimization function and any constraints
ourselves from a written description.
