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Makovka662 [10]
3 years ago
5

The number of people , d, in thousands applying for medical benefits per week in a particular city can be modules by the equatio

n d(t)=2.5 sin(0.76t+0.3)+3.8 where t is the time in years since January 2000. Based on the equation The Maximum number of people in thousands apply for benefits per year in the city is
Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
3 0

Answer:

<h2>4773 peoples.</h2>

Step-by-step explanation:

Given the number of people d, in thousands applying for medical benefits per week in a particular city c modeled by the equation d(t)=2.5 sin(0.76t+0.3)+3.8 where t is the time in years, the maximum number of people tat will apply will occur at d(t)/dt = 0

Differentiating the function given with respect to t, we will have;

d(t)=2.5 sin(0.76t+0.3)+3.8

First we need to know that differential of any constant is zero.

Using\ chain\ rule\\\frac{d(t)}{dt} = 2.5cos(0.76t+0.3) * 0.76 + 0\\ \\\frac{d(t)}{dt} = 1.9cos(0.76t+0.3)

If \frac{d(t)}{dt} =0 then;

1.9cos(0.76t+0.3) = 0\\\\cos(0.76t+0.3)  = 0\\\\0.76t+0.3  = cos^{-1} 0\\\\0.76+3t = 90\\\\3t = 90-0.76\\3t = 89.24\\\\t = 89.24/3\\\\t = 29.75years

To know the maximum number of people in thousands that apply for benefits per year in the city, we wil substitute the value of t = 29.75 into the modeled equation

d(t)=2.5 sin(0.76t+0.3)+3.8\\d(29.75) = 2.5 sin(0.76(29.75)+0.3)+3.8\\d(29.75) = 2.5 sin(22.61+0.3)+3.8\\\\d(29.75) = 2.5 sin(22.91)+3.8\\\\d(29.75) = 0.9732+3.8\\d(29.75) = 4.7732\\\\

Since d is in thousands, the maximum number of people in thousands will be 4.7732*1000 = 4773.2 which is approximately 4773 peoples.

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