MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 11 18.01 Fall 2006 Lecture 11: Max/Min Problems Example 1. y = ln x (same function as in last lecture) x x0=e1/eFigure 1: Graph of y = ln x . x 1What is the maximum value? Answer: y = .• e • Where (or at what point) is the maximum achieved? Answer: x = e. (See Fig. 1).) Beware: Some people will ask “What is the maximum?”. The answer is not e. You will get so used to finding the critical point x = e, the main calculus step, that you will forget to find the maximum 1 1value y = . Both the critical point x = e and critical value y = are important. Together, they e e1form the point of the graph (e, ) where it turns around. e Example 2. Find the max and the min of the function in Fig. 2 Answer: If you’ve already graphed the function, it’s obvious where the maximum and minimum values are. The point is to find the maximum and minimum without sketching the whole graph. Idea: Look for the max and min among the critical points and endpoints.You can see from Fig. 2 that we only need to compare the heights or y-values corresponding to endpoints and critical points. (Watch out for discontinuities!) 1maxminFigure 2: Search for max and min among critical points and endpoints Example 3. Find the open-topped can with the least surface area enclosing a fixed volume, V. rhFigure 3: Open-topped can. 1. Draw the picture. 2. Figure out what variables to use. (In this case, r, h, V and surface area, S.) 3. Figure out what the constraints are in the problem, and express them using a formula. In this example, the constraint is V = πr2h = constant We’re also looking for the surface area. So we need the formula for that, too: S = πr2 + (2πr)h Now, in symbols, the problem is to minimize S with V constant. 2 Lecture 11 18.01 Fall 2006� � 4. Use the constraint equation to express everything in terms of r (and the constant V ). h = V ; S = πr2 + (2πr) V 2πr πr2 5. Find the critical points (solve dS/dr = 0), as well as the endpoints. S will achieve its max and min at one of these places. dS 2V 3 V � V �1/3 dr = 2πr − r2 = 0 = ⇒ πr3 − V = 0 = ⇒ r = π = ⇒ r = π We’re not done yet. We’ve still got to evaluate S at the endpoints: r = 0 and “r = ∞”. 2V S = πr2 + , 0 ≤ r < ∞r 2As r → 0, the second term, r , goes to infinity, so S → ∞. As r → ∞, the first term πr2 goes to infinity, so S → ∞. Since S = +∞ at each end, the minimum is achieved at the critical point r = (V/π)1/3, not at either endpoint. srto ∞to ∞Figure 4: Graph of S We’re still not done. We want to find the minimum value of the surface area, S, and the values of h. � �1/3 � �−2/3 � �1/3V V V V V V r = π ; h = πr2 = π � V �2/3 = π π = π π � �2/3 � �1/3 S = πr2 + 2 V = πV + 2VV = 3π−1/3V 2/3 r π π Finally, another, often better, way of answering that question is to find the proportions of the can. In other words, what is h r ? Answer: h r = (V/π)1/3 (V/π)1/3 = 1. 3 Lecture 11 18.01 Fall 2006Example 4. Consider a wire of length 1, cut into two pieces. Bend each piece into a square. We want to figure out where to cut the wire in order to enclose as much area in the two squares as possible. (1/4)x0x1(1/4)(1-x)Figure 5: Illustration for Example 5. 2 x xThe first square will have sides of length . Its area will be . The second square will have � �24 16 sides of length 1−4 x . Its area will be 1−4 x . The total area is then � x �2 � 1 − x �2 A = +4 4 A� = 216 x + 2(116− x)(−1) = x 8 − 18 + x 8= 0 = ⇒ 2x − 1 = 0 = ⇒ x = 12 So, one extreme value of the area is � 1 �2 � 1 �2 1 A = 2 + 2 = 4 4 32 We’re not done yet, though. We still need to check the endpoints! At x = 0, A = 02 + � 1 − 0 �2 =1 4 16 At x = 1, � �21 1 A = + 02 = 4 16 4 Lecture 11 18.01 Fall 2006By checking the endpoints in Fig. 6, we see that the minimum area was achieved at x = 12 . The maximum area is not achieved in 0 < x < 1, but it is achieved at x = 0 or 1. The maximum corresponds to using the whole length of wire for one square. 1/211/161/32xAreaFigure 6: Graph of the area function. Moral: Don’t forget endpoints. If you only look at critical points you may find the worst answer, rather than the best one. 5 Lecture 11 18.01 Fall
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