University:
Glendale Community College
Course:
MAT155 | College Algebra/Functions with Review
Academic year:

2020
Views:

63

Pages:

2
Author:

Bella Cruz

Chapter 4 - handout 3 1) L + 2w = 250 A = lw = (250 - 2w)w A = 250w - 2w² A' = 250 - 4w A' = 0 → w = 62.5 A'' = -4 < 0 ABS max y = 54 when x = 9 ABS min y = -250 when x = 5 2) f(x) = x³ - 75x f'(x) = 3x² - 75 3(x² - 25) = 0 (x + 5)(x - 5) = 0 x = -5, 5 3) f(x) = 7xe^x f'(x) = 7e^x + 7xe^x 7e^x(1 + x) = 0 x = -1 4a) A = lw = (250 - 2w)w A = 250w - 2w² A' = 250 - 4w A'' = -4 < 0 ABS max y = 54 when x = 9 ABS min y = -250 when x = 5 4b) A = lw = (250 - 2w)w A = 250w - 2w² A' = 250 - 4w A'' = -4 < 0 ABS max y = 54 when x = 9 ABS min y = -250 when x = 5 4c) 0 ≤ w ≤ 125 A = 250w - 2w² A' = 250 - 4w A'' = -4 < 0 ABS max y = 54 when x = 9 ABS min y = -250 when x = 5 4d) A = 250w - 2w² A' = 250 - 4w A'' = -4 < 0 ABS max y = 54 when x = 9 ABS min y = -250 when x = 5 4e) L = 250 - 2w L = 250 - 2(62.5) = 125 m A = lw A = (125)(62.5) A = 7812.5 m² 5) dy/dx = 4x - 1 d²y/dx² = 4 > 0 ABS min y = -7/e when x = -1 8) Given: dr/dt = 2 r = 5 Find: dA/dt Solution: A = πr² dA/dt = 2πr * dr/dt dA/dt = 2π(5)(2) dA/dt = 20π ft²/sec Area increasing 20π ft²/sec 9) Given: dr/dt = 4 r = 10 Find: dA/dt Solution: dA/dt = 2πr * dr/dt dA/dt = 2π(10)(4) dA/dt = 80π ft²/sec Area increasing 80π ft²/sec 10) Given: dv/dt = 4 r = 8 Find: dr/dt Solution: V = (4/3)πr³ dV/dt = 3 * (4/3)πr² * dr/dt 4 = 4πr² * dr/dt 4 = 4π(8)² * dr/dt 1/(64π) = dr/dt dr/dt = 1/(64π) cm/sec Radius growing 1/(64π) cm/sec 11) Given: dv/dt = 7 r = 6 Find: dr/dt Solution: dV/dt = 4πr² * dr/dt 7 = 4π(6)² * dr/dt 7/(144π) = dr/dt dr/dt = 7/(144π) cm/sec Radius growing 7/(144π) cm/sec