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“Applications of Derivatives” Handout
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Related Rates
General steps:
1. Draw a picture!! (may not be possible for every problem, but usually is)
2. Label everything. If a quantity is fixed for the entire problem, write in the number. If it can change, then
assign it a variable. There are often multiple ways to draw and label things, but the final answer will be the
same.
3. Write down what you know, and what you want to know. Note: When writing down given/known rates of
change, make them positive if the variable is getting larger, negative if the variable is getting smaller (this is
going to depend on how you labeled your picture).
4. Figure out how everything is related and come up with some formula relating the variables. This can involve
using geometric formulas, triangles, similar triangles, etc.
5. Differentiation implicitly with respect to t. Remember, in these problems all variables are viewed as a function
d
of t, so you’ll pick up dt
terms from the chain rule.
6. Plug in known values for variables and rates, then solve for the quantity in which you’re interested. Warning:
DO NOT plug in numbers for things that can change until after you differentiate!! If a quantity is constant
throughout the entire problem and cannot change, then you should have already put in the picture as a number
in step (2) . Replacing a variable with an expression involving another variable is allowed.
Note: In the book, they make any answers that are rates positive, even if the quantity is decreasing (the negative is
assumed from the word “decreasing”). In class I’ve kept the negative on the answer to emphasize the fact that the
quantity is decreasing, and on the test I would want to see the negative.
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Derivatives and Max/Min Values, and the Shapes of Curves
Definition 2.1. Let f (x) be defined on an interval I containing c.
• f (c) is said to be the absolute maximum of f on I if and only if f (c) ≥ f (x) for all x in I.
• f (c) is said to be the absolute minimum of f on I if and only if f (c) ≤ f (x) for all x in I.
Theorem 2.2 (Extreme Value Theorem). If f is continuous on a closed interval [a, b], then f has an absolute
maximum and an absolute minimum on [a, b].
Definition 2.3. Let f (x) be defined on an interval I containing c.
• f (c) is said to be a local maximum (or relative maximum) of f on I if and only if f (c) ≥ f (x) for all x “near
c”.
• f (c) is said to be a local minimum (or relative minimum) of f on I if and only if f (c) ≤ f (x) for all x “near c”.
Definition 2.4. Let f (x) be a function defined on an interval I.
• f (x) is said to be increasing on I if f (x1 ) ≤ f (x2 ) for any two numbers x1 , x2 in I such that x1 < x2 .
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(a) f (x) is increasing
(b) f (x) is increasing
(c) f (x) is decreasing
(d) f (x) is decreasing
Figure 1: Some Examples of Increasing and Decreasing Functions
• f (x) is said to be decreasing on I if f (x1 ) ≥ f (x2 ) for any two numbers x1 , x2 in I such that x1 < x2 .
Definition 2.5. A critical value (or critical number ) x = c of a function f (x) is any number [in the domain of f ]
such that f ′ (c) = 0 or f ′ (c) is undefined.
Definition 2.6 (formal definition of concavity). A function f is said to be concave up on an interval I if f ′ is an
increasing function on I.
f is said to be concave down on an interval I if f ′ is an decreasing function on I.
Definition 2.7 (informal definition of concavity). “Concave up, like a cup. Concave down, like a frown.”
Definition 2.8. A point where f changes from concave up to concave down or from concave down to concave up is
called an inflection points or a point of inflection.
(a) A concave up function
(b) A concave down function
Figure 2: Concavity Examples
(a) Increasing and Concave Up (b) Increasing and Concave Down (c) Decreasing and Concave Up (d) Decreasing and Concave Down
Theorem 2.9 (Fermat’s Theorem). If f has a local max/min and f ′ (c) is defined, then f ′ (c) = 0. More generally,
if f has a local max/min at c, then c is a critical number of f .
Theorem 2.10 (“Increasing/Decreasing Theorem”). Let f (x) be a differentiable function on some interval I,
1. If f ′ (x) > 0 on I, then f (x) is increasing on I.
2. If f ′ (x) < 0 on I, then f (x) is decreasing on I.
Theorem 2.11 (The First Derivative Test). Suppose c is a critical number of a [continuous] function f ,
1. If f ′ changes from (+) → (−) (i.e., it goes from increasing to decreasing) at c, then f has a local max at
(c, f (c))
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2. If f ′ changes from (−) → (+) (i.e., it goes from decreasing to increasing) at c, then f has a local min at
(c, f (c))
3. If f ′ does not change sign (i.e., have (+) → (+) or (−) → (−)) or if f (c) is undefined, then f has no local
max/min at c.
Theorem 2.12 (The Second Derivative Test). Suppose f ′′ (x) is continuous near c,
1. If f ′ (c) = 0 and f ′′ (c) > 0, then f has a local minimum at c.
2. If f ′ (c) = 0 and f ′′ (c) < 0, then f has a local maximum at c.
3. If f ′′ (c) = 0, or if f ′ (c) or f ′′ (c) is undefined, then this test is inconclusive.
Theorem 2.13 (“Concavity Test”, this is what some people mean by “the second derivative test”). If f ′′ (x) > 0
for all x in an interval I, then f (x) is concave up on I. If f ′′ (x) < 0 for all x in an interval I, then f (x) is concave
down on I.
Theorem 2.14 (Mean Value Theorem). If f is differentiable on the closed interval [a, b], then there exists a number
c between a and b (i.e., a 6 c 6 b) such that
f ′ (c) =
f (x) − f (a)
x−a
This is saying that at some point in the interval, the instantaneous rate of change, f ′ (c), is equal to the average rate
(a)
of the change over the entire interval, f (x)−f
.
x−a
How we use all this stuff:
Finding the absolute maximum and absolute minimum of a continuous function y = f (x) on a closed
interval [a, b]:
0. Check to make sure the interval is closed (easy) and that the function is continuous on the interval (it’s okay
to have discontinuities outside the interval).
1. Find all critical values of f in the interval (we throw out any that aren’t in the interval).
2. Evaluate f (not f ′ ) at those critical points and at a and b (the endpoints of the interval).
3. The largest value from step 2 is the absolute maximum of f on [a, b], the smallest value is the absolute minimum
of f on [a, b].
Notes: When finding abs max/min, we’re interested in the function values (y’s); for many other problems in this
chapter, we’re more interested in the x-values. The abs max or min may occur at more than one x-value.
Finding intervals on which a function is increasing, and where it’s decreasing, and finding local maxima
and local minima:
1. Find all critical values (x-values where f ′ = 0 or is undefined). . .
2. . . . and plot them on a number line - I’ll use a filled in circle • at critical values where f is defined, and an open
circle ◦ where f is not defined. This divides the number line into intervals.
3. Pick a test value from each interval (I’ll denote these with an x) and plug into f ′ to see if the derivative is (+)
or (−) (we don’t need the actual value of f ′ at each test value - only its sign)
4. Interpret:
– f is increasing on the (+) intervals.
– f is decreasing on the (−) intervals.
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– If f ′ changes from (+) → (−) at a critical value and f is defined there (i.e., (+) → (−) at •), then we have
a local maximum at that x-value (plug back into the original function to get the y-value).
– If f ′ changes from (−) → (+) at a critical value and f is defined there (i.e., (−) → (+) at •), then we have
a local minimum at that x-value (plug back into the original function to get the y-value).
– If f ′ doesn’t change or if f is undefined at a critical value (i.e., (−) → (−), (+) → (+), or ◦), then we have
neither a local max nor a local min there.
Note: If f ′ doesn’t change at a critical value and the function is defined there, then we can combine those two
intervals and that critical value into one interval.
Finding intervals on which a function is concave up, and where it’s concave down, and finding inflection
points:
(Since concavity is defined in terms of f ′ being increasing or decreasing, these steps will be similar to the ”find
inc/dec” steps.)
1. Find all points where f ′′ = 0 or is undefined (there are the critical values of f ′ ). . .
2. . . . and plot them on a number line. This divides the number line into intervals. I’ll use:
• a filled in circle • at values where both f ′ and f are defined,
◦ an open circle ◦ at values where both f ′ and f are undefined,
•
•
◦ and a filled circle over an open circle ◦ at values where f ′ is undefined but f is defined.
3. Pick a test value from each interval (I’ll denote these with an x) and plug into f ′′ to see if the second derivative
is (+) or (−) (we don’t need the actual value of f ′′ at each test value - only its sign)
4. Interpret:
• f is concave up on the (+) intervals.
• f is concave down on the (−) intervals.
• If f ′′ changes from (+) → (−) or from (−) → (+) at a value and f is defined there (i.e., (+) → (−) or
•
(−) → (+) at • or ◦), then we have an inflection point at that x-value (plug back into the original function
to get the y-value).
• If f ′′ doesn’t change or if f is undefined at a value (i.e., (−) → (−), (+) → (+), or ◦), then we don’t have
an inflection point there.
Note: If f ′′ doesn’t change at a value and f ′ is defined there (i.e., (−) → (−) or (+) → (+) at •), then we can
combine those two intervals and that value into one interval.
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3
Optimization Problems
General steps:
1. Draw a picture and assign variables.
2. Write down the equation to be maximized or minimized (this is sometimes called the objective equation) and
the equation that describes the constraint (this is sometimes called the constraint equation).
3. Use the constraint equation to rewrite the objective equation so that it has only one independent variable.
4. Find the domain of the [new] objective equation.
5. Find the maximum or minimum using calculus. Technically we’re finding an absolute max/min, but in these
problems it very often occurs at a local max/min in the domain (and in many problems there’s only one
possibility).
6. Verify your answer using the 2nd Derivative Test.
Exercises
1. A farmer wants to build a pen with two dividers in order to separate elephants, donkeys, and penguins. If 600
ft of fence is available and one side of the pen is bounded by a river (and needs no fence since all the animals
just happen to have an irrational fear of water), then what is the maximum area that can be enclosed?
2. A carpenter wants to build a rectangular box with square sides. The material for the bottom costs $20/ft2 ,
material for the sides costs $10/ft2 , and the material for the top costs $50/ft2 t. If the volume of the box must
be 5 ft3 , then find the dimensions that will minimize the cost (and find the minimum cost).
3. A right circular cylinder is inscribed in a sphere of radius 6 in. Find the largest possible volume of such a
cylinder.
4. A peach orchard owner wants to maximize the amount of peaches produced by her orchard. She has found
that the per-tree yield is equal to 900 whenever she plants 45 or fewer trees per acre, and that when more than
45 trees are planted per acre, the per-tree yield decreases by 25 peaches per tree for every extra tree planted.
For example, if there were 40 trees planted per acre, each tree would produce 900 peaches. If there were 50
trees planted per acre, each tree would produce 900 − 25(50 − 45) = 775 peaches. Find the value of x that
maximizes the yield and the maximum value of the yield.
5. A 5 in × 8 in piece of paper has a square cut out of each corner (same size from each) and is then folded to
make an open-top box. Find the size of the square that will maximize the volume.
6. Find
√ √the area of the largest rectangle that can be inscribed inside an isosceles triangle with side lengths
2, 2, 2.
7. A box with a square base and open top must have a volume of 32000 cm3 . Find the dimensions of the box
that will minimize the amount of material needed.
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8. Cheerwine⃝
Bottling Company has decided to make a foray into alcoholic beverages with the introduction of
a new drink: Cheerbeer. They have asked you to design a new cylindrical can that will hold 355 ml and uses
the least amount of aluminum. What should be the dimensions of the can?
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Newton’s Method
To approximate a root/zero of a function, make an initial “guess” (doesn’t have to be a good one) about where it is,
call it x0 , and then apply the recursion:
f (xold )
xnew = xold − ′
f (xold )
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until the output repeats out to however many decimal places you want. This is the approximate location of a root.
√
Notes: This recursion doesn’t always converge (try f (x) = 3 x and any x0 other than 0), and if there are multiple roots then you will need to use different x0 ’s to find each one.
Figure 3: Four iterations of Newton’s method applied to f (x) = ex−3 − 4 with x0 = 6.9
How to do this in your calculator:
The nicer way:
1. Go under your graphing or function screen and
input the algorithm as a function (remember the
name): x − f (x)/f ′ (x). Be sure it’s doing the subtraction and division correctly (use parenthesis as
needed).
2. Return to the main screen, input x0 , and press
ENTER
3. Evaluate the function from the first step at ANS. For
example, if your function was stored as y1, then you
would need to type in y1(ANS). Press ENTER. The result is x1 .
The less convenient way:
1. Input x0 and press ENTER
2. Type in the algorithm for your specific function, using ANS as your variable: ANS − f (ANS)/f ′ (ANS)
3. Press ENTER. The result is x1 .
4. Pressing ENTER repeatedly will give x2 , x3 , . . ., etc.
Continue until these numbers agree out to the desired number of decimal places.
4. Pressing ENTER repeatedly will give x2 , x3 , . . ., etc.
Continue until these numbers agree out to the deNote: This method isn’t as convenient as the first since
sired number of decimal places.
you have to re-input the function each time you want to
Note: The main challenge with this method is calling the try a different x0 or make a mistake.
function correctly. This method makes it much easier to
run the algorithm again for different x0 ’s.
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