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

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

Question
Functions
asked 2021-02-03
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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