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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3 years ago