Expert answers to "Let P(t)=100+20cos⁡⁡6t,0≤t≤π2. Find the maximum and minimum values..."

Dottie Parra

Answered question

2020-10-26

Let

P (t) = 100 + 20 \cos 6 t, 0 \leq t \leq \frac{π}{2}

. Find the maximum and minimum values for P, if any.

Answer & Explanation

Talisha

Skilled2020-10-27Added 93 answers

$P (t) = 100 + 20 \cos 6 t, 0 \leq t \leq \frac{π}{20} \leq t \leq \frac{π}{2}$

Differentiating P(t) with respect to t :
$P' (t) = - 20 \times 6 \times \sin 6 t = - 120 \sin 6 t$
$P' (t) = 0 \geq \sin 6 t = 0$
$\sin ⊘ = 0 \geq ⊘ = n \times π$
$\sin 6 t = 0 \geq t = n \frac{π}{6}$
Given

$0 \leq t \leq π / 20 \leq t \frac{π}{2}$
So

$t = 0, \frac{π}{2}, \frac{π}{6} t = 0, \frac{π}{2}, \frac{π}{6}$
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

$t = \frac{π}{2} t = \frac{π}{2}$

$P (\frac{π}{2}) = 100 + 20 \cos (6 \times \frac{π}{2})$
$P (\frac{π}{2}) = 100 + 20 \cos 3 π$
$P \frac{π}{2} = 100 - 20$
$P \frac{π}{2} = 80$
When

$t = \frac{π}{6} t = \frac{π}{6}$
$P \frac{π}{6} = 100 + 20 \cos 6 \times \frac{π}{6}$
$P \frac{π}{6} = 100 - 20 P \frac{π}{6} = 100 - 20$
$P \frac{π}{6} = 80 P \frac{π}{6} = 80$
Thus we get maximum value of P(t) at t=0 as 120
Thus we get minimum value of P(t) at $t = \frac{π}{2}$ and $\frac{π}{2}$ and $\frac{π}{6}$ as 80

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