Math 231E. Fall 2013. Lecture 5C.Max/MinFrom here on out, I will stand for an interval. It can be open, closed, or semi-open.Definition 2. The absolute maximum of f on I is the largest value f takes on I. Specifically,we say that M is the absolute maximum of f if there is an x with f(x) = M, and, for any y ∈ I,f(y) ≤ M.Note! it must be a value of f(x). The function f(x) = 2x + 1 on the interval (3, 5) has noabsolute maximum! Also, the function f(x) = 1/x on (0, ∞) also has no maximum.Definition 3. A local maximum is a value f(c) such that f(c) ≥ f(x) for all x near c.(The same definitions work for absolute and local minimum in the obvious manner.)Theorem 9 (Extreme Value Theorem). If f(x) is continuous on [a, b], then f attains an absolutemaximum and an absolute minimum in [a, b].Theorem 10 (Fermat’s (not last) Theorem). If f(x) has a local maximum, or local minimum, atc, and f0(c) exists, then f0(c) = 0.Proof. Say f(x) has a local maximum at c (the proof is similar for a minimum). For h > 0 smallenough, we havef(c + h) ≤ f(c).This meansf(c + h) − f(c) ≤ 0,f(c + h) − f(c)h≤ 0,limh→0+f(c + h) − f(c)h≤ 0.We also have for h < 0 and small enough, that f(c + h) ≤ f (c), orf(c + h) − f(c) ≤ 0,f(c + h) − f(c)h≥ 0,limh→0−f(c + h) − f(c)h≥ 0.(Note! Why did the inequality flip??) But the limit only exists if both the left-hand and right-handlimits are the same, so this limit must be zero.Definition 4. A critical point is a number c such that f0(c) = 0 or f0(c) does not exist.Algorithm to find absolute max/min• Find critical points of f,• Evaluate f at all of these points,• Evaluate f at the endpoints,• the absolute max and min must be one of these.Let us work a few examples:Example 24. Given a fixed perimeter, how do we maximize and minimize the area of a rectangle?Let us consider a fixed perimeter of 100 m. We have 2x + 2y = 100 m, and the area is A = xy.Let us solve for y and get y = 50 m − x, soA(x) = x(50 m −x) = 50 mx − x2.The domain of validity for this function is clearly 0 m ≤ x ≤ 50 m, since we cannot allow x or y tobe negative. We then check for critical points: A0(x) = 50 m −2x, so setting this equal to zero givesx = 25 m. Thus we have to check three points; the critical point and the two boundary points:A(0 m) = 0 m2, A(25 m) = (25 m)(25 m) = 625 m2, A(50 m) = 0 m2.Clearly the interior point is an absolute maximum, and the two boundary points are absoluteminima. Thus we have proved that the optimal rectangle is a square; the pessimal rectangle is thedegenerate one where we choose one side length or the other to be zero.We can generalize this to an arbitrary perimeter L. If we form a rectangle with width x andheight y, then the perimeter is L = 2x + 2y, while the area is A = xy. Since L is the constraint,let us write y = L/2 − x, and then the area becomesA(x) = xL2− x=L2x − x2.The domain of validity of this function is 0 ≤ x ≤ L/2 (if x > L/2 then y would be negative!).Again we findA0(x) =L2− 2x,and thus our critical point is x = L/4. Again, we check the boundary points, and we obtainA(0) = 0, A(L/4) = (L/4)(L/4) =L216, A(L/2) =
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