Question

The rate at which a machine​ operator's efficiency, E​ (expressed as a​ percentage), changes with respect to time t is given by ​, where t is the number of hours the operator has been at work. a​) Find​ E(t), given that the​ operator's efficiency after working 2 hr is ​%; that​ is, ​E(2). b​) Use the answer in part​ (a) to find the​ operator's efficiency after 3​ hr; after 5 hr.

Answer 1

a) The operator's efficiency function is
`\( E(t) = 25t - 4t^2 + 33 \)` after working 2 hours.

b) After 3 hours, efficiency is 72%, and after 5 hours, efficiency is 58%, derived from the calculated function.

a) Find E(t), given that the operator's efficiency after working 2 hr is 67% (E(2) = 67):

Given the differential equation
`\( (dE)/(dt) = 25 - 8t \)`, we need to solve it to find
`\( E(t) \).`

`\[ \int dE = \int (25 - 8t)dt \]`

Integrating both sides:

`\[ E(t) = \int (25 - 8t)dt \]`

\[ E(t) = 25t - 4t^2 + C \]

Now, use the given condition E(2) = 67 to solve for C:

`\[ E(2) = 25(2) - 4(2)^2 + C = 67 \]`

50 - 16 + C = 67

C = 33

So, the operator's efficiency function is
`\( E(t) = 25t - 4t^2 + 33 \).`

b) Use the answer in part (a) to find the operator's efficiency after 3 hr and after 5 hr:

`\[ E(3) = 25(3) - 4(3)^2 + 33 \]`

E(3) = 75 - 36 + 33

E(3) = 72

After 3 hours, the operator's efficiency is 72%.

`\[ E(5) = 25(5) - 4(5)^2 + 33 \]`

E(5) = 125 - 100 + 33

E(5) = 58

After 5 hours, the operator's efficiency is 58%.

The probable question may be:

The rate at which a machine​ operator's efficiency, E​ (expressed as a​ percentage), changes with respect to time t is given by dE/dt= 25-8t​, where t is the number of hours the operator has been at work. a​) Find​ E(t), given that the​ operator's efficiency after working 2 hr is ​67%; that​ is, ​E(2)=67. b​) Use the answer in part​ (a) to find the​ operator's efficiency after 3​ hr; after 5 hr.