3.7 Optimization Problems
Guidelines for solving Optimization Problems:
#1. Read the problem until you understand it
a) note what is given
b) what is the unknown quantity to be optimized
#2. Draw a picture – label any parts that are important to the problem
#3. Introduce Variables
a) List using relation in the picture and in the problem as an equation or algebraic
expression.
b) Identify the unknown variable
#4. Write an equation for the unknown quantity ---- (Primary Equation)
a) express the unknown as a function of a single variable or in 2 equations in 2
unknowns. (2 equations ---- will require manipulation)
#5. Find the critical points and then test critical points and endpoints of domain in the
unknowns.
Ex. #1 p. 285
What is the smallest perimeter possible for a rectangle whose area is 16 in². What are the
dimensions of the rectangle?
--Use A = l x w
L x w = 16
and P = 2l + 2w
w = 16/l
P = 2l + 2(16/l) ----- 2l + 32/l
** Now find the derivative = 2 - 32/l²
** get l² out of denominator by dividing everything by l² == 2l² - 32
** Set = 0 to find critical points
2l² - 32 = 0
2(l – 4)(l + 4) = 0
L = -4, 4
** must be > 0 because dealing with dimensions so l=4
16 = 4 x w so w = 4 and P = 2(4) + 2(4) = 16
Ex. A manufacturer wants to design an open box having a square base and a surface area of 108
square inches. What dimensions will produce a box with maximum volume?
Volume = x²h (lxwxh)
Surface Area = area of base + area of all 4 sides
108 = x² + 4xh
Because volume is maximized we want to express V as a function of just 1 variable so solve for
h:
x² + 4xh = 108
h = (108 - x²)
4x
V = x² (108-x²)
4x
** now plug h back into original equation
= 108x² - x4
4x
= 27x - x³
4
Now to find maximum value of V – determine a feasible domain – where do the values of x
make sense in this problem?
**know V≥0 and that x can’t be - and that Area of base = x² is at most 108
So feasible Domain = 0≤x≤√108
**to maximize V - find critical numbers so differentiate with respect to x
Dy/dx 27x - x³
= 27 – 3x²
4
4
** now set =0 and find critical pts.
27 – 3x² = 0
4
x²=36
x = -6, 6 (only want +6)
**Now evaluate endpoints and critical points
V(0) = 0 (0,0)
V(6) = 162 – 54 = 108 (6,108) ** so x = 6. h = 3 and x = 6
V(√108) = 0 (√108, 0)
Economics Optimization:
Marginal Revenue dr/dx
Marginal Cost dc/dx
Marginal Profit dp/dx
r(x) = revenue for selling x items
c(x) = cost of producing x items
p(x) = r(x) – c(x) profit from producing and selling x items
if p(x) = r(x) – c(x) has a max value it occurs at production level at which p′(x) = 0
** at a production level yielding max profit – marginal revenue = marginal cost
Ex #43.
You offer a tour service that offers the following:
*$200 per person if 50 people (the min. to book a tour) go on the tour
* For each additional person, up to a max of 80 people total, the rate per person is
reduced by $2
It costs $6000 (a fixed cost) plus $32 per person to conduct the tour. How many people does it
take to maximize your profit?
Let x = the # of people
(50 + x) = the # of people over 50
(200 – 2x) = reduced rate for each person if over 50 book
Cost = 32(50 + x) – 6000
P(x) = r(x) – c(x) so p(x) = (50+x)(200-2x)-32(50 + x) -6000
P(x) = -2x² + 68x + 2400
** Now take derivative and set = 0 to find critical points
-4x + 68 = 0 x = 17
So (50 + 17) would maximize profit = 67 people
4.7 Indeterminate Forms and L’Hopital’s Rule
Suppose we are trying to analyze the behavior of the function f(x) = lnx
x -1
** although f is not defined at x=1, we need to know how f(x) behaves near 1
(in other words – we want to know the value of the limit)
** In computing the limit – we see that the Numerator and Denom. approach 0
and are not defined this leads to:
Indeterminate Form (0/0 form) – occurs when continuous functions f(x) and g(x) are both 0 at
x=a, so lim f(x) cannot be found by substituting x=a, it would result in 0/0
g(x)
x--a
**in our study of limits we found that it we used cancellation, rearrangement of
Terms, or some other algebraic manipulations, it often led to indeterminate 0/0
Form.
** However, if we used the form lim f(x) – f(a) the long form of the derivative
x–a
x—a
it would prevent getting the indeterminate form. This leads to:
L’Hopital’s Rule (First Form)
Suppose that f(a)=g(a) = 0, that f′(a) and g′(a) exist and that g′(a) ≠0 then
Lim f(x) = f′(a)
g′(a)
** treat as 2 separate functions
x ---a g(x)
(show graphical representation)
Ex. a) lim 3x – sinx
x
x---0
b) lim √1 + x - 1
x—0
x
** sometimes, though, after differentiation, the new Numerator and Denominator both still = 0 at
x=a, then we use the stronger form of L’Hopital’s Rule
L’Hopital’s Rule (Stronger Form)
Suppose that f(a) = g(a)=0 and that f and g are differentiable in the open interval, I, containing a
and that g′(x)≠a on I if x≠a then
Lim f(x) = lim f′(x)
x—a g(x)
x—a g′(x)
**basically just keep deriving til no 0 in denomin.
Ex. a) lim √1 + x -1 + x/2
x—0
x²
b) lim x – sinx
x—0
x³
**L’Hopital’s Rule does not apply when either the Numerator or Denominator has a finite
nonzero limit --- Must be indeterminate form of 0/0
Ex. lim 1-cosx
x—0 x + x²
L’Hopital’s Rule with One Sided Limits:
Ex. lim
sin x
+
x—0
x²
lim
sin x
x—0
x²
Indeterminate Forms ∞/∞ - if in the form where both f(x)---∞/-∞ and g(x) --- ∞/-∞, then lim
f′(x) may or may not exist provided that the limit on the right exists
x—a g′(x)
Ex. lim
sec x
x--π/2 1+tan x
b) lim
x- 2x²
x--∞ 3x² + 5x
c) lim ex
x--∞ x²
Indeterminate Form ∞∙0
If lim f(x) = 0 and lim g(x) = ∞/-∞ then it is not clear what the values of lim f(x)g(x)
x—a
x—a
x—a
if any, will be – there becomes a struggle between f and g.
a) if f wins – then the answer = 0
b) if g wins – then the answer may be ∞ or -∞
c) **may be a compromise where the answer is a finite nonzero #
** You deal with this by writing f(x)g(x) as a quotient: fg = f or fg = g
1/g
1/f
(put whichever = 0 in numerator)
Ex. Lim (xsin 1 )
x--∞
x
Steps: 1 evaluate f and g at x--∞
Step 2 Then derive each the n and d separately
Step 3 Solve
Indeterminate Difference
Ex. Lim ( 1 - 1 )
x—0 sinx x
if x—0+ then sin x—0+ and 1/sinx – 1/x = ∞ - ∞
if x—0- then sin x—0- and 1/sinx – 1/x = -∞- -∞=-∞+∞
Steps: to find out what happens to the limit – combine functions:
So:
4.7 Newton’s Method – a technique to approximate the solution to an equation f(x) =0
** It uses tangent lines in place of the graph of y=f(x) near the points where f is 0
GOAL: for estimating a solution of an equation f(x) = 0 is
a) To produce a sequence of approximations that approach the solution
--starts with an initial guess (x0) and improves the guess 1-step at a time.
Refer to graph:
Procedure for Newton’s Method: - let f(c)=0 where f is differentiable on an open interval
Containing c, then to approximate c do the following:
#1. Make an initial estimate, x0 that is close to c
#2. Use the 1st approximation to get a 2nd, then use the 2nd to get a 3rd, and so on using:
Xn+1 = xn - f(xn) if f′(xn) ≠0
f′(xn)
Ex. f(x) = x²-2 we use x0 = 1 as initial guess (graph in calculator)
a) f(x) =x² - 2
f′(x) = 2x
so: using above formula: xn+1 = xn - xn² - 2
2xn
= xn+1 = xn – xn + 1
2 xn
X0=1 so plug in 1 1 - ½ + 1/1 = 1.5
X1 = 1.5
1.5 – .75 + (1/1.5) = 1.41667
X2 = 1.41667
= 1.414216
X3= 1.414216
** if we keep carrying this procedure out you will see it converges
To 1.4142…. where each recursion results in more decimals places
Carried out, but with the same first 5 decimals – approaches a limit
(we know this particular answer is √2 = 1.414213562
Purpose of Newton’s method – it is used to calculate roots b/c they converge so fast.
Ex. Find the x-coordinate of point where curve y = x³ - x crosses horizontal line at y=1
Start by writing f(x) = x³ -x = 1 then solve by getting = to 0
Convergence of Newton’s Method: as in above examples, the approximation approaches
a limit – the sequence x1, x2, x3,…xn is said to converge – therefore
if the limit is c, then it can be shown that c must be a 0 of f.
When Newton’s Method doesn’t yield a convergent sequence:
#1. B/C Newton’s method involves division by f′(xn) – it is clear that the method will fail if the
derivative is 0 for any xn in the sequence
a) Correct this by = overcome it by choosing a different starting value of x
#2 When successive approximations go back and forth between 2 values (iterative formula.
Conditions for Convergence:
│f(x)f′′(x)│ < 1
│[f′(x)]² │
Ex. f(x) = x²-2
f′(x) = 2x f′′(x) = 2
= │(x²-2)(2)│ = │ 1 - 1 │
│[2x]² │
│ 2 x² │
What about on interval (1,3) - just plug into above and see if <1 and it will indicate
convergence.
#3. When Newton’s method converges to a root, it may not be the root you have in mind.
Fractal Basins and Newton’s Method
a) finding roots by Newton’s method can be uncertain in the sense that for some equation --- the final outcome can be extremely sensitive to the starting values location
Ex. f(x) = 4x4 – 4x² (look at figure 4.52 in book p. 304
 there are 3 roots you can see from graph
 since there are so many roots you have “basins of attraction”
Fractal Basins – when each root has infinitely many “basins’ of attraction in the complex plane –
each basin has a boundary where complicated patterns repeat without end under successive
manipulations.
Exercise #6 Use Newton’s method to find the negative fourth root of by solving the equation x4
– 2 = 0. Start with x0 = 1 and find x2