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lesya692 [45]
3 years ago
15

Mr. Johnson buys 5 bottles of lemonade for the school picnic. He buys four 28

Mathematics
1 answer:
DochEvi [55]3 years ago
7 0
(4x28)+64
112+64
176
Rounded to the nearest ten it would be 180
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Answer:

The absolute maximum is \frac{3\sqrt 3}2 and the absolute minimum value is 0.

Step-by-step explanation:

Differentiate of f both sides w.r.t.  t,

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Now take f'(t)=0

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\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0

\Rightarrow (\sin t+1)(2\sin t-1)=0

\Rightarrow \sin t+1=0  \;\text{and}\; 2\sin t-1=0

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In the interval 0\leq t\leq \frac {\pi}2, the answer to this problem is \frac {\pi}6

Now find the second derivative of f(t) w.r.t.   t,

f''(t)=-2\cos t-4\sin 2t

\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3

Thus, f(t) is maximum at t=\frac {\pi}6 and minimum at t=0

\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0

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