(A) surely you can do this. They gave you the rate as an equation. So,
dx/dt at t=1 is
30(1-(20/((1+20)^2))) = 28.64 widgets/week
(B) the total widgets during the first n weeks is just the sum of all the widgets produced during each week. That is, the area under the dx/dt curve. That is, the integral. So, for the first four weeks (up to the beginning of week 5), the total production is
∫[0,4] 30(1-(20/((t+20)^2))) dt = 115
(C) ∫[0,3] ...
When you come across a problem that seems totally unsolvable, you have to remind yourself that is must have something to do with what you have just covered. So, go back to look for similar problems in the text. There will always be a clue there. They won't just be asking random questions that have nothing to do with what you have been taught!!
I was given a homework assignment that I need to do but have no idea where to even start. If possible, please include steps. Thank You!!!
A factory manufactures widgets. The rate of production of widgets after t weeks is
(dx/dt)=30(1-(20/((t+20)^2))) widgets/week.
A) Determine the rate at which widgets are produced at the start of week 1. Set up the expression but do not simplify.
B) Determine the number of widgets produced from the beginning of production to the beginning of the fifth week.
C) Determine the number of widgets produced from the beginning of production to the beginning of the fourth week.
2 answers
Actually, I see that I misread (A). At the start of week 1 is after ZERO weeks, so at t=0,
dx/dt = 30(1-(20/((0+20)^2))) = 30(1 - 1/20) = ...
dx/dt = 30(1-(20/((0+20)^2))) = 30(1 - 1/20) = ...
4 answers