University:
Whitman College
Course:
MATH-125 | Calculus I
Academic year:

2024
Views:

397

Pages:

1
Author:

utupimiem4a

M125 Fall 2003 Quiz 4 SOLUTIONS Show all your work! 1. Find the maximum of f (x) = 4x + x9 , for 1 ≤ x ≤ 3. First we find the critical points of f , which are where the derivative is zero or does not exist: f 0 (x) = 4 − 9 x2 The critical points are x = 0 (outside the interval) and: 4= of which only x = 3 2 9 9 3 ⇒ 4x2 = 9 ⇒ x2 = ⇒ x = ± x2 4 2 is inside the interval. Now we have to check the value of f at the endpoints and critical points. At x = 1, f (1) = 4 + At x = 3 2, 9 1 f (3/2) = 4 · = 13 3 2 At x = 3, f (3) = 4 · 3 + + 9 3 9 3/2 =2·3+9· 2 3 = 6 + 6 = 12 = 12 + 3 = 15 Therefore, the maximum value of f is 15 at x = 3. 2. Find the maximum of ab2 if a, b must be positive and a + b = 5. We have a choice between substitution for a or b. It’s easier to substitute for a: a=5−b So f (b) = (5 − b)b2 = 5b2 − b3 . Now find the critical points: f 0 (b) = 0 ⇒ 10b − 3b2 = 0 ⇒ b(10 − 3b) = 0 so that b = 0 or b = 10/3. The values of b can range between 0 and 5, so we can check these values as well; f (0) = f (5) = 0. 5 100 500 f (10/3) = · = 3 9 27 3. Extra Credit: In the last problem, what is the answer if 5 is replaced by an arbitrary positive constant k? In this case, f (b) = (k − b)b2 = kb2 − b3 so the critical points are: f 0 (b) = 2kb − 3b2 = b(2k − 3b) = 0 so b = 0 or b = 2k 3 . In the first case, the product is zero, and if b = k, the area is also zero. Therefore, if f (2k/3) > 0, it will be the max: f (2k/3) = k− 2k 3 2 2k k 4k 2 4k 3 = · = · 3 3 9 27 which is positive since k > 0. 1

Calculus I Quiz 4 Solutions

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