Transcript Aim: How can we apply the first derivative to solve easy

Aim: How can we apply the first derivative to
solve easy, medium and hard level problems
related to Optimization?
Donow: Try these compositions of
functions
If f(X)=2, f’(3)=-1, g(3)=3,
g’(3)=0, Find F’(3)
(question borrowed
from Mrs. Dela Cruz)
a.
F(x)= 2f(x)- g(x)
Just think here to your
self, is there any
product rule or
division rule
involved?
(con).
No right? Good. Let’s simply make the f(x)
and g(x) into f’(x) and g’(x) respectively.
(dont’ worry about the two!!!)
 2f’(x)-g’(x) = F’(x)
Now just plug in:
F’(3)= 2f’(3)- g’(3)
Based of our charts we see that f’(3)= -1
g’(3)= 0

Continued…
So, its F’(3)= 2(-1)-0= -2
That’s it, see how easy it is…

Let’s try another one:
F(x)= 1/2f(x)g(x)
Ask yourself: Does this question involve the
product rule or the division rule Yes!!!!
The Product Rule
Well, don’t panic! There is a simple
product rule we can apply:
 F’(x)= f’g+g’f.
You may say: Here we go again…
Well, this derivative is simply the derivative
of f times g added to the derivative of g
mulitplied with the f value.
Don’t get it? Well, lets approach the
example:

Solving: ½ f(x)g(x)…
Problem
F’(x)=
1/2[f(x)g’(x)+g(x)f’(x)
F’(3) =
1/2[f’(3)+g(3)+g’(3)f(
3)

F’(3)= ½[2(0)+3(-1)]
The algorithm
Application of the
product rule
Placing what needs to
be derived, in the
equation
Plugging in with the
values of the original
equation

Continuation of the donow prob.

F’(3)= -3/2

Evaluating the
expression.
Why the donow problem?
You see what I was trying to show you?
Tell me, what was common about both
things: The first derivative.
For all of them, I was trying to show you
that the first derivative plays an integral
role in our finding of the derivatives.
In our lesson, it will play a bigger role
Now on to our lesson…

What is Optimization?
Optimization is basically trying to get the
most of something while getting rid of the
least. It can also be trying to get least
amount of something.
You see the big oil companies: Exxon Mobil,
Chevron Corporation and BP trying to
make the most amount of money, but
trying to cut their costs in a highly
competitive environment.
Optimization functions (con)

Sometimes, we want to get the least
volume with the most amount of material,
so that we don’t waste any.
As you can see, Optimization has a wide
array of applications from big name gas
tankers to simple cubic volumes.
I’ll help you dissect the problems…

Question 1
Intimidated? Relax, we’ll start with
something easy:
(borrowed from www.math.ucdavis.edu)
Find two nonnegative numbers whose sum
is 9 and so that the product of one
number and the square of the other
number is a maximum.
Approaching the problem
You may have difficulty if you don’t break
the problem down into simple parts:
The problem states: “nonnegative numbers”
So I tell myself “Only positive numbers”
Next, it says “ sum of 9” We don’t know
what the sum of the problem, is so we
just have to make them up.
Approaching the problem…con.
Let’s say that “x” is the first number, and “y
is the second number. So, it follows that
x+y=9. Simple enough?
Ok, take a deep breath…
Now, read the second part of the problem…
(the product of one number and the square
of the other is a maximum).
Seems a little tricky, right?
Continued.
We know that one number is “x” and the other
number is the “square,” So, this means that the
“y” is the y squared.
Let P represent the product of the equation:
P  xy
2
Now, we must attempt to connect both parts,
which is probably the toughest part, that I had
trouble with…
The connection:
Well, we want to find the
product of the
equation,so our goal is to
make both of the
variables in the product
the same( You can select
either variable you want
to put in terms of, but
lets use “y”, for simplicity
It follows that:

y  9 x
P  x(9  x) 2
Derivative Time:)

Remember what I had told you before about the
first derivative, well now it’s time to apply it…

P’ = f’g+g’f
P' = x (2) ( 9-x)(-1) + (1) ( 9-x)2
= ( 9-x) [ -2x + ( 9-x) ]
= ( 9-x) [ 9-3x ]
= ( 9-x) (3)[ 3-x ]




=0
 X=3 or 9.

Plugging back…
Remember the plugging back, now we do
the same…
If we do 9*0 or 0*9, well have 0. So the
only ones that work is 3 and 6, which well
get 108 as the maximum product.
Example 2:
Borrowed from
http://www.math.ucdavis.edu Medium
Problem:
A sheet of cardboard 3 ft. by 4 ft. will be
made into a box by cutting equal-sized
squares from each corner and folding up
the four edges. What will be the
dimensions of the box with largest volume
?
Solving the problem
Drawing a picture is the safest way to start,
draw a picture, and label the sides!
Other steps:

Here is the Rectangular Box without the
cover:
Steps:

Observe that the problems that we solve are
similar to that of the previous problem. These
are: writing the equation, getting the “solved
for” equation in terms of the same variable.
Finding the first derivative, setting it equal to 0,
and then solving for the variable. If we want to
check whether the answer is maximum, we plug
our answer into the second dx/dy. If the answer
is less than 0, were okay, otherwise, it does not
exist.
Finally solved…





V =LWH = (4-2x) (3-2x) (x) .
V' = (-2) (3-2x) (x) + (4-2x) (-2) (x) + (4-2x) (3-2x) (1)
= -6x + 4x2 - 8x + 4x2 + 4x2 - 14x + 12
= 12x2 - 28x + 12
or
.
= 4 ( 3x2 - 7x + 3 )

= 0 (notice we had to use the product rule three times,
since there are 3 “parts

Now, you must use the quadratic quation to solve for the
x value. You will find that the answer comes out to x=
0.57 or 1.77. Now, you must plug back in to the original
equation. You will find that the largest volume is 3.03
cubic feet at x=0.57.
Hard problem
Is it getting easier? I certainly hope so. Now
will come the hardest one of all…
Borrowed from
http://www.math.ucdavis.edu
Consider a rectangle of perimeter 12 inches.
Form a cylinder by revolving this rectangle
about one of its edges. What dimensions
of the rectangle will result in a cylinder of
maximum volume ?
Approaching the toughie…

Let’s draw a picture, we know that the
rectangle is inside the cylinder, and only
half way. So here it is.
Dissecting the problem
We know that the perimeter is the sum of twice the
width and twice the length.
So, its 12 = 2r + 2h , now, we need to find the
equation of the volume of the cylinder. Let p
represent pi.
V=pr2h. Now, notice that we have “h in both
equations. Since we have this, we can make the
perimeter equation in terms of r.
Now, if we do that, all we have to do, is follow the
aforementioned rules of finding the first derivative,
setting it equal to 0, and then plugging in. We
should then have the maximum volume.

Solved answer.
V’=p(12r-3r2 )
p(3r)(4-r)
3pr(4-r)=0
r= 0 or 4

Now, you have to guess different values.
You'll find that when r=4 and when h=2,
there is the highest volume. This volume
will be 100 feet cubed.