Math 1300 4.2 Optimization, part 2 Name: Solutions 1. Consider the ...

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Math 1300 4.2 Optimization, part 2 Name: Solutions2. Consider the function f(x) = 2x 3 − 6x 2 − 144x.(a) Find the global maximum and minimum values of f(x) on [−6, 6].f ′ (x) = 6x 2 − 12x − 144 = 6(x 2 − 2x − 24) = 6(x − 6)(x + 4) so our critical pointsare x = −4, 6.f(−4) = 352 f(6) = −648 f(−6) = 216So the global max is 352 and occurs at x = −4 and the global min is −648 and occursat x = 6.(b) Find the global maximum and minimum values of f(x) on (−5, 5).From our critical points −4 and 6 we now only care about −4.f(−4) = 352lim f(x) = f(−5) = 320x→−5f(x) = f(5) = −620limx→5So f has no global maximum or global minimum on (−5, 5).
Math 1300 4.2 Optimization, part 2 Name: Solutions3. Consider the function f(x) = 5 ln(x) − x.(a) Find the global maximum and minimum values of f(x) on [3, 10].f ′ (x) = 5 x − 10 = 5 − 1 ⇒ x = 5 is our critical point.xf(5) = 5 ln(5)−5 ≈ 3.047 f(3) = 5 ln(3)−3 ≈ 2.493 f(10) = 5 ln(10)−10 ≈ 1.513So the global max is 5 ln(5) − 5 and occurs at x = 5, and the global min is 5 ln(10) − 10and occurs at x = 10.(b) Find the global maximum and minimum values of f(x) on (0, ∞).As before, our only critical point is at x = 5, and f(5) = 5 ln(5) − 5 ≈ 3.047.limx→0x→0f(x) = lim (5 ln(x) − x) = −∞+ +lim f(x) = lim (5 ln(x) − x) = −∞ (by looking at the graph)x→∞ x→∞So f has a global max value of 5 ln(5) − 5 at x = 5, but no global minimum.
Page 1: Math 1300 4.2 Optimization, part 2

Math 1300 4.2 Optimization, part 2 Name: Solutions 1. Consider the ...

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