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Vini *BANNED* · #1 10-28-2010, 04:22 AM

The efficiency coefficient (E) of a team measured related to is number of team members (t) can be described by the equation E(t) = 2000 + 108t - 4t^3 T € [1,7].

a) Use calculus to determine the number of team members that produces greatest efficiency

b) What is the maximum efficiency coefficient that can be achieved by the team?

#2 10-28-2010, 04:54 AM

a) Take the derivative of this function to get 108-12*(t^2). Set this equal to 0, to get t^2=9. t then must be 3.

b) Now plug in 3 into the original equation. You should get 2216

Vini *BANNED* · #3 10-28-2010, 05:06 AM

Quote:

Originally Posted by video

a) Take the derivative of this function to get 108-12*(t^2). Set this equal to 0, to get t^2=9. t then must be 3.

b) Now plug in 3 into the original equation. You should get 2216

you did it so easy, i had to do 108t - 12t^2
= 12(9-t^2)
= 12(3- t)(3-t)
t= 3, t = -3

but idk how you got t^2=9.

#4 10-28-2010, 05:13 AM

Quote:

Originally Posted by Vini

you did it so easy, i had to do 108t - 12t^2
= 12(9-t^2)
= 12(3- t)(3-t)
t= 3, t = -3

but idk how you got t^2=9.

108-12t^2=0
12t^2=108
t^2=9

Does that work? We also know that t can't be a negative number

Vini *BANNED* · #5 10-28-2010, 05:28 AM

How about this one:

The hardness (H) of a temporary glue, h hours after application is given by:

H = 2000 + (H - 7)^3

a) Determine the rate of change of H (dH/dh)

b) what is the rate of change of H when h= 0?

c) When is the glue at its hardest?

#6 10-28-2010, 06:07 AM

Quote:

Originally Posted by Vini

How about this one:

The hardness (H) of a temporary glue, h hours after application is given by:

H = 2000 + (H - 7)^3

a) Determine the rate of change of H (dH/dh)

b) what is the rate of change of H when h= 0?

c) When is the glue at its hardest?

I think you mean H=2000+(h-7)^2 (lowercase h)

a) dH/dh=3(h-7)^2

b) dH/dh (0)=3(0-7)^2=49*3=147

c) Set dH/dh=0. h=7