Question

find the maximum value of 2sin theta+3​

Answer 1

Answer:

Calculate the value of maxima and minima of given question-

y(max)=2sin

2

θ−3sinθ+2

Differentiate both side w.r.t

dx

dy

=4sinθcosθ−3sinθ

Hence-

dθ

dy

=0

4sinθcosθ−3cosθ=0

Take common cosθ we get:

cosθ(4sinθ−3)=0

Now we get

cosθ=0

θ=90

∘

(4sinθ−3)=0

sinθ=

4

3

θ=48.59

Again, differentiating

dθ

dy

, we get

dθ

2

d

2

y

=4(cos

θ

−sin

2

θ)+3sinθ

Evaluate the value of maxima and minima after putting the value of θ=90

∘

, we get:

Case 1:-

θ=90

∘

=4cos180

0

+3sin90

∘

dθ

2

d

2

y

<−1

Y is max. Value when θ=90

∘

Case 2:-

θ=48.59

∘

=4cos97.18

0

+3sin48.59

∘

dθ

2

d

2

y

>0

Y is minimum value when θ=48.59

∘

θ=48.59

∘

So putting the value of θ=90

∘

y(max)=2sin

2

(−90)

∘

−3sin(−90)

∘

+2

=2+3+2

=7

Putting the value [θ=48.59

∘

]

y(min)=2sin

2

48.59

∘

−3sin48.59

∘

+2

=7/8

Hence, there is an answer..