Question

Alabama Instruments Company has set up a production line to manufacture a new calculator. The rate of production of these calculators after t weeks is dx dt = 5000 1 − 100 (t + 10)2 calculators/week. (Notice that production approaches 5000 per week as time goes on, but the initial production is lower because of the workers' unfamiliarity with the new techniques.) Find the number of calculators produced from the beginning of the third week to the end of the fourth week. (Round your answer to the nearest calculator.)

Answer 1

Answer:

Calculators from the beginning of the third week to the end of the fourth week = 4048.

Step-by-step explanation:

We know that the rate of production of these calculators after t weeks is given by

$\frac{dx}{dt} =5000(1-\frac{100}{(t+10)^{2}})$

To find the number of calculators that have been produced in a period, we need to take the integral of the function above; the desired time is t=2 (beginning of third week) to t=4 (end of the fourth week). Therefore, the number of calculators produced in the given time is

$\int\limits^4_2 {\frac{dx}{dt} } \, dt = \int\limits^4_2 {5000(1-\frac{100}{(t+10)^{2} }) } \, dt$

Substitute $t+10=u$ and $dt=du$, observe that the limits of integration will change

$\int\limits^4_2 {\frac{dx}{dt} } \, dt => \int\limits^{14}_{12} {\frac{du}{dt} } \, dt$

$5000\int\limits^{14}_{12} { 1-\frac{100}{u^{2} } } \, du$

$5000(u+100u^{-1})\left \{ {{14} \atop {12}}\right.\\5000(2+\frac{100}{14}-\frac{100}{12} )\\4047.62$ ≈ 4048

Answer 2

The number of calculators produced from the beginning of the third week to the end of the fourth week is; 4048

How to solve first order differential equations?

We are given the rate of production of the calculators after t weeks as;

dx/dt = 5000[1 − (100/(t + 10)²)] calculators/week

To find the number of calculators that have been produced in a period, we will integrate the above function between t = 2 to t=4.

Thus, the number of calculators produced in the given time is;

$\int\limits^4_2 {\frac{dx}{dt}} \, dt = \int\limits^4_2 {5000(1 - \frac{100}{(t + 10)^{2} } } \, dt$

If we put t + 10 for u and du for dt, it means that our limits of integration will be; u = 14 and 12. Thus;

$5000\int\limits^{14}_{12} {1 - \frac{100}{u^{2} } } \, du$

Integrating this with a calculator gives 4048 calculators

Read more on differential equations at; https://brainly.com/question/18760518