hw 20 - absolute extrema section 3.6: problem 10\n(1 point)\nan employees monthly productivity m, in number…

hw 20 - absolute extrema section 3.6: problem 10\n(1 point)\nan employees monthly productivity m, in number of units produced, is found to be a function of the number t of years of service. for a certain product, a productivity function is shown below. find the maximum productivity and the year in which it is achieved.\n$m(t)=-8t^{2}+344t + 180$ for $0\\leq t\\leq43$\nabsolute maximum value:\n(round to three decimal places as needed.)

hw 20 - absolute extrema section 3.6: problem 10\n(1 point)\nan employees monthly productivity m, in number of units produced, is found to be a function of the number t of years of service. for a certain product, a productivity function is shown below. find the maximum productivity and the year in which it is achieved.\n$m(t)=-8t^{2}+344t + 180$ for $0\\leq t\\leq43$\nabsolute maximum value:\n(round to three decimal places as needed.)

Answer

Explanation:

Step1: Find the derivative of (M(t))

The derivative of (M(t)=-8t^{2}+344t + 180) is (M^{\prime}(t)=-16t + 344) using the power rule ((x^{n})^\prime=nx^{n - 1}).

Step2: Set the derivative equal to zero and solve for (t)

Set (M^{\prime}(t)=0), so (-16t+344 = 0). [ \begin{align*} -16t&=-344\ t&=\frac{344}{16}=21.5 \end{align*} ]

Step3: Check the endpoints and the critical point

  • When (t = 0), (M(0)=-8(0)^{2}+344(0)+180 = 180)
  • When (t = 21.5), (M(21.5)=-8(21.5)^{2}+344(21.5)+180) [ \begin{align*} M(21.5)&=-8\times462.25+344\times21.5 + 180\ &=-3698+7496+180\ &=3978 \end{align*} ]
  • When (t = 43), (M(43)=-8(43)^{2}+344(43)+180) [ \begin{align*} M(43)&=-8\times1849+344\times43+180\ &=-14792+14792+180\ &=180 \end{align*} ]

Answer:

(3978.000)