Question

the largest value of 5sin theta+9 is ___.​

Answer 1

Given :  5sinθ + 9

To Find : largest value  

Solution

5sinθ + 9

sinθ range is [-1 , 1]

maximum value of sinθ = 1

Hence 5sinθ + 9

= 5(1) + 9

= 14

Largest Value is 14

largest value of 5sinθ+9 is 14

f(θ) = 5sinθ + 9

f'(θ) = 5cosθ

5cosθ  = 0

=> θ = π/2 , 3π/2

f''(θ) = -5Sinθ

f''  (π/2) = -5Sinπ/2 = - 5 < 0  Hence maximum value at  π/2

f''  (3π/2) = -5Sin3π/2 = 5 > 0  Hence minimum value at  3π/2

f(π/2) = 5sinπ/2 + 9  = 5 + 9 = 14     largest value

f(3π/2) = 5sinπ/2 + 9  = -5 + 9 =  4   Smallest Value

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Answer 2

SOLUTION

TO DETERMINE

The largest value of

$\sf{5 \sin \theta \: + 9}$

EVALUATION

SOLVE USING CONCEPT OF DIFFERENTIATION

Let

$\sf{y = 5 \sin \theta \: + 9}$

Differentiating both sides with respect θ to we get

$\displaystyle \sf{ \frac{dy}{d \theta} = 5 \cos \theta }$

Again Differentiating both sides with respect θ to we get

$\displaystyle \sf{ \frac{ {d}^{2} y}{d { \theta}^{2} } = - 5 \sin \theta }$

For maximum value of y we have

$\displaystyle \sf{ \frac{dy}{d \theta} = 0 }$

$\implies \displaystyle \sf{ 5 \cos \theta = 0}$

$\implies \displaystyle \sf{ \theta = \frac{\pi}{2} \: , \: \frac{3\pi}{2} }$

Now

$\displaystyle \sf{ \frac{ {d}^{2} y}{d { \theta}^{2} } \bigg| _{ \theta = \frac{\pi}{2} } = - 5 < 0}$

$\displaystyle \sf{Hence \:y \: has \: maximum \: value \: at \: \theta = \frac{\pi}{2} \: }$

$\displaystyle \sf{ \frac{ {d}^{2} y}{d { \theta}^{2} } \bigg| _{ \theta = \frac{3\pi}{2} } = 5 > 0}$

$\displaystyle \sf{Hence \:y \: has \: minimum \: value \: at \: \theta = \frac{3\pi}{2} \: }$

Hence the largest ( maximum) value

$\displaystyle \sf{ y \bigg| _{ \theta = \frac{\pi}{2} } = 5 \sin \: \frac{\pi}{2} + 9}$

$= 5 + 9$

$= 14$

FINAL ANSWER

The largest value of $\sf{(5 \sin \theta \: + 9}) \: \: \: is \: \: \: 14$

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