Question 83.
Using calculus find the maximum value of 6sin (x + 2π/7) - 8sin (x - 3π/14).
Also find the maximum value of
5sin (x+π/30) + 2√2 sin(x-π/20) + √3sin (x+π/5) + sin (x+11π/30) + 2sin (x+8π/15) + 2√3sin (x+7π/10).
Answer 83.
f(θ) = 6sinθ + 8cosθ
For f(θ) to be maximum, f '(θ) = 0 and f "(θ) < 0
f '(θ) = 0 => 6cosθ - 8sinθ = 0 => θ = kπ + arctan(3/4)
f "(θ) = - (6sinθ + 8cosθ) < 0 for θ = arctan(3/4)
=> sinθ = 3/5 and cosθ = 4/5
=> maximum value of f(θ) = 6*(3/5) + 8*(4/5) = 10.
5sin (x+π/30) + 2√2sin (x-π/20) + √3sin (x+π/5) + sin (x+11π/30)
+ 2sin (x+8π/15) + 2√3sin (x+7π/10)
= 5sin(x + 6°) + 2√2 sin(x - 9°) + √3 sin(x + 36°) + sin(x + 66°)
+ 2sin(x + 96°) + 2√3sin(x + 126°)
= 5siny + 2√2 sin(y -15°) + √3 sin(y + 30°) + sin(y + 60°)
+ 2sin(y + 90°) + 2√3sin(y + 90° + 30°) [Taking x+6°=y]
= 5siny + 2√2(sinycos15° - cosysin15°)
+ √3(sinycos30° + cosysin30°) + (sinycos60° + cosysin60°)
+ 2cosy + 2√3(cosycos30° - sinysin30°)
= 5siny + 2√2[(√6+√2)/4*siny - (√6-√2)/4*cosy]
+ √3(√3/2*siny + 1/2*cosy) + (1/2*siny + √3/2*cosy)
+ 2cosy + 2√3(√3/2*cosy - 1/2*siny)
= (5+√3+1+3/2+1/2-√3)siny + (1-√3+√3/2+√3/2+2+3)cosy
= 8siny + 6cosy
Maximum value of this is 10.
LINK to YA!
Using calculus find the maximum value of 6sin (x + 2π/7) - 8sin (x - 3π/14).
Also find the maximum value of
5sin (x+π/30) + 2√2 sin(x-π/20) + √3sin (x+π/5) + sin (x+11π/30) + 2sin (x+8π/15) + 2√3sin (x+7π/10).
Answer 83.
f(θ) = 6sinθ + 8cosθ
For f(θ) to be maximum, f '(θ) = 0 and f "(θ) < 0
f '(θ) = 0 => 6cosθ - 8sinθ = 0 => θ = kπ + arctan(3/4)
f "(θ) = - (6sinθ + 8cosθ) < 0 for θ = arctan(3/4)
=> sinθ = 3/5 and cosθ = 4/5
=> maximum value of f(θ) = 6*(3/5) + 8*(4/5) = 10.
5sin (x+π/30) + 2√2sin (x-π/20) + √3sin (x+π/5) + sin (x+11π/30)
+ 2sin (x+8π/15) + 2√3sin (x+7π/10)
= 5sin(x + 6°) + 2√2 sin(x - 9°) + √3 sin(x + 36°) + sin(x + 66°)
+ 2sin(x + 96°) + 2√3sin(x + 126°)
= 5siny + 2√2 sin(y -15°) + √3 sin(y + 30°) + sin(y + 60°)
+ 2sin(y + 90°) + 2√3sin(y + 90° + 30°) [Taking x+6°=y]
= 5siny + 2√2(sinycos15° - cosysin15°)
+ √3(sinycos30° + cosysin30°) + (sinycos60° + cosysin60°)
+ 2cosy + 2√3(cosycos30° - sinysin30°)
= 5siny + 2√2[(√6+√2)/4*siny - (√6-√2)/4*cosy]
+ √3(√3/2*siny + 1/2*cosy) + (1/2*siny + √3/2*cosy)
+ 2cosy + 2√3(√3/2*cosy - 1/2*siny)
= (5+√3+1+3/2+1/2-√3)siny + (1-√3+√3/2+√3/2+2+3)cosy
= 8siny + 6cosy
Maximum value of this is 10.
LINK to YA!
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