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# Let P(t)=100+20 cos⁡ 6t,0leq tleq frac{pi}{2}. Find the maximum and minimum values for P, if any.

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Functions
asked 2020-10-26
Let $$P(t)=100+20 \cos⁡ 6t,0\leq t\leq \frac{\pi}{2}$$. Find the maximum and minimum values for P, if any.

## Answers (1)

2020-10-27
$$P(t)=100+20\cos6t , 0\leq t\leq \frac{\pi}{20}\leq t\leq\frac{π}{2}$$ Differentiating P(t) with respect to t :
$$P′(t)=−20\times 6\times \sin 6t=−120\sin 6t$$
$$P′(t)=0 \geq \sin6t=0$$
$$\sin\oslash =0 \geq \oslash=n\times \pi$$
$$\sin6t=0\geq t=n\frac{\pi}{6}$$
Given 0\leq t\leq \pi /20\leq t\frac{\pi}{2}
So t=0,\frac{\pi}{2},\frac{\pi}{6}t=0,\frac{\pi}{2},\frac{\pi}{6}
On putting the value of t in P(t)we get: When t=0
$$P(0)=100+20\cos (6\times 0)$$
$$P(0)=100+20\times 1$$
$$P(0)=120$$
When t=\frac{\pi}{2} t=\frac{\pi}{2} $$P(\frac{\pi}{2})=100+20\cos(6\times \frac{\pi}{2})$$
$$P(\frac{\pi}{2})=100+20\cos 3\pi$$
$$P\frac{\pi}{2}=100−20$$
$$P\frac{\pi}{2}=80$$
When t=\frac{\pi}{6}t=\frac{\pi}{6}
$$P\frac{\pi}{6}=100+20\cos 6\times \frac{\pi}{6}$$
$$P\frac{\pi}{6}=100−20P\frac{\pi}{6}=100−20$$
$$P\frac{\pi}{6}=80P\frac{\pi}{6}=80$$
Thus we get maximum value of P(t) at t=0 as 120
Thus we get minimum value of P(t) at t=\frac{\pi}{2} and \frac{\pi}{2} and \frac{\pi}{6} as 80

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