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

Question
Functions
Let $$\displaystyle{P}{\left({t}\right)}={100}+{20}{\cos{ ## Answers (1) 2021-02-04 \(\displaystyle{P}{\left({t}\right)}={100}+{20}{\cos{{6}}}{t},{0}\leq{t}\leq{\frac{{\pi}}{{{20}}}}\leq{t}\leq{\frac{{π}}{{{2}}}}$$ Differentiating P(t) with respect to t :
$$\displaystyle{P}′{\left({t}\right)}=−{20}\times{6}\times{\sin{{6}}}{t}=−{120}{\sin{{6}}}{t}$$
$$\displaystyle{P}′{\left({t}\right)}={0}\geq{\sin{{6}}}{t}={0}$$
$$\displaystyle{\sin{{o}}}{s}{l}{a}{s}{h}={0}\geq{o}{s}{l}{a}{s}{h}={n}\times\pi$$
$$\displaystyle{\sin{{6}}}{t}={0}\geq{t}={n}{\frac{{\pi}}{{{6}}}}$$
Given 0\leq t\leq \pi /20\leq t\frac{\pi}{2}
So t=0,\frac{\pi}{2},\frac{\pi}{6}t=0,\frac{\pi}{2},\frac{\pi}{6}
On putting the value of t in P(t)we get: When t=0
$$\displaystyle{P}{\left({0}\right)}={100}+{20}{\cos{{\left({6}\times{0}\right)}}}$$
$$\displaystyle{P}{\left({0}\right)}={100}+{20}\times{1}$$
$$\displaystyle{P}{\left({0}\right)}={120}$$
When t=\frac{\pi}{2} t=\frac{\pi}{2} $$\displaystyle{P}{\left({\frac{{\pi}}{{{2}}}}\right)}={100}+{20}{\cos{{\left({6}\times{\frac{{\pi}}{{{2}}}}\right)}}}$$
$$\displaystyle{P}{\left({\frac{{\pi}}{{{2}}}}\right)}={100}+{20}{\cos{{3}}}\pi$$
$$\displaystyle{P}{\frac{{\pi}}{{{2}}}}={100}−{20}$$
$$\displaystyle{P}{\frac{{\pi}}{{{2}}}}={80}$$
When t=\frac{\pi}{6}t=\frac{\pi}{6}
$$\displaystyle{P}{\frac{{\pi}}{{{6}}}}={100}+{20}{\cos{{6}}}\times{\frac{{\pi}}{{{6}}}}$$
$$\displaystyle{P}{\frac{{\pi}}{{{6}}}}={100}−{20}{P}{\frac{{\pi}}{{{6}}}}={100}−{20}$$
$$\displaystyle{P}{\frac{{\pi}}{{{6}}}}={80}{P}{\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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