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Section 3.3 Maximum and Minimum Val-
ues
Definition For a function f defined on a set S of real numbers and a number c in S.
A) f (c) is called the absolute maximum of f on S if f (c) ≥ f (x) for all x in S.
B) f (c) is called the absolute minimum of f on S if f (c) ≤ f (x) for all x in S.
Absolute maximum and minimum are also called absolute extrema. Simi- larly we may also define the local extrema as follows.
Definition Let I be an open interval containing the point c
a) f (c) is called a local maximum of f, if f (c) ≥ f (x) for all x in I
b) f (c) is called a local minimum of f, if f (c) ≤ f (x) for all x in I.
How are the definitions of absolute and relative extrema different? In the relative case, we only require that our point f(c) beat values f(x) where x is close to c, though in the absolute case we insist that f(c) beat all values f(x) for x in the domain of f. Lets see the difference in action in the example below:
Example In the graph below the function is defined on the interval I = [− 5 , 7]. And f has absolute maximum at x = −5 and a absolute min- imum at x = 7. It has local maximum at x = 2 and local minimum at x = − 3 , x = 0.
Example In the graph below the function is defined on the interval [0, 2]. And f has an absolute maximum at x = 2 and and absolute minimum at x = 0. It has a local minimum at x = 1 and no local maximum.
Cautionary Example 1This graph gives an example of a continuous func- tion on the open interval (0, 1) which does not satisfy the conclusion of the Extreme value theorem. Hence no surprise that we didn’t get the absolute max or min the theorem promises.
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Cautionary Example 2 This second graph belongs to a function f (x) de- fined on a closed interval which again neither has an absolute minimum nor an absolute maximum. (Notice that this function fails to be continuous on the interval [0, 1] hence why the theorem cannot be used)
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Before we can list the ways how we will find the absolute extremum we need one definition and two theorems.
Definition A number c in the domain of f is called a critical number(value) of f if f ′(c) = 0 or f ′(c) not defined.
Fermat’s Theorem Suppose that f (c) is a local extremum. Then c must be a critical number of f.
Theorem Suppose that f is continuous on the closed interval [a, b]. Then the absolute extrema of f occurs at an endpoint (a or b) or at a critical number.
For the proof of the last two theorems please check out your books. The proofs given are quite easily understandable. I’ll be happy to answer any questions you have about them outside of the class.
The last theorem gives us the following recipe to find absolute minimum and maximum of a function on a closed interval.
Finding Absolute Maxima and Minima
(i) Verify that f (x) is continuous on a closed interval of interest.
(ii) Find critical points of f (x) that lie in the closed interval of interest
(iii) Evaluate f (x) at critical points and endpoints in a chart
(iv) Pick out the absolute maximum by finding which critical point/endpoint gives the largest function value (similarly for absolute minimum)
Remark It is IMPORTANT if you’re going to use this process that you verify the first step. If you don’t first check that the function is continuous on a closed interval, this method can fail! Now let’s apply it :
Example Find the absolute maxima and minima of the function f (x) = x
x − x^2 on the interval [0, 1]
We want to use the procedure above, so first we need to verify that f(x) is continuous on [0, 1]. But since x is continuous everywhere and
x − x^2 is defined on the interval [0, 1] (and hence continuous there) we see that f(x) is continuous on [0, 1] as desired.
For the second step, we need to find critical points of f(x). This means we need to compute f ′(x). Now
x f(x) -1 f (−1) = 1 0 f (0) = 0 2 f (2) = 2 Since 0 is the smallest value we see for f (x) in this chart, we have that x = 0 is the absolute minimum of f(x). Since 2 is the largest value this means x = 2 is an absolute maximum of f(x).
ExampleA crazy billionaire gives you 10 meters of gold wire and asks you to construct a rectangle with maximum area. If you succeed, he’ll give you and your math professor $1,000,000. What rectangle will you construct? What will be its dimensions and area?
Let’s assume that you will want to win the money, and hence you’ll try to maximize the area of the resultant rectangle. How will you do this? Let’s call x the length of the rectangle and y its width.
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x
y
A=xy
One can see that the resultant rectangle has area A = xy. We would like to maximize this function, but we don’t know how to in its current form: it’s a function of two variables, and we don’t know how to maximize such a function. So what do we do?
The fact that we have 10 meters of wire comes to the rescue, since it tells us the perimeter of our figure is 10. Specifically, we have 10 = 2x + 2y. We can now solve for y in terms of x and rewrite our area function: 10 = 2x + 2y ⇒ y = 5 − x
and hence the area A(x) = xy = x(5 − x) = 5x − x^2 is the kind of function we can maximize, but first we need to know the domain of the function. For
this, notice that we could make a rectangle with no height (so that x = 0). This would be dumb, but we could do it. We could make another dumb rectangle: one without width, so that x = 5 (think about why this is 5 and not 10). The possible rectangles we could construct sit somewhere be- tween these two stupid extremes, so we see that x lives in the interval [0, 5]. Now we are really in business since we have reduced our original problem to the following: Maximize the function A(x) = 5x−x^2 on the interval [0, 5].
Since A(x) is continuous on [0, 5] (it’s a polynomial, so in fact its continuous everywhere), we can find the extreme values of A by computing the value of A(x) at critical points and endpoints. We proceed to find critical points: A′(x) = 5 − 2 x, and so the only critical point we have is x = 52. Now we need to evaluate A(x) at the critical point and the endpoints of the interval.
x f(x) 0 A(0) = 0 5 2 A(^
5 2 ) =^
25 4 5 A(5) = 0 So to maximize the area of our rectangle we will make x = 52 (and so y = (^52) since y = 5−x), and the resultant area will be 254. We can restate our result: of all rectangles with a fixed perimeter, a square maximizes area.
