Question

This is my last problem:) Find the time t in years when the annual sales x of a new product are increasing at the greatest rate. Use a graphing utility to verify your results. Please help:) Will give medal

Answer 1

\[x=\frac{ 600000t^2 }{ 36+ t^2 }\]

Answer 2

@jim_thompson5910

Answer 3

are you in calculus?

Answer 4

Not sure, but I can't reply to messages =/
I'm in Business Calc

Answer 5

I'm not sure how to start the problem

Answer 6

are you able to find the derivative of that function?

Answer 7

I believe is (36+t^2)(1200000)-(2T)(T^2

Answer 8

I may be completely wrong

Answer 9

close,

(36+t^2)(1,200,000t) - (600000t^2)(2t)

all over (36+t^2)^2

Answer 10

this function represents how much x is changing with respect to t

we want to find the greatest rate of change, so we have to derive again because we want to locate the local max of this function

Answer 11

to help make things a bit simpler, the derivative simplifies to

(43200000t)/(36+t^2)^2

Answer 12

now you have to derive (43200000t)/(36+t^2)^2

Answer 13

this part I believe is where I got stuck

Answer 14

what do you get when you derive (43200000t)/(36+t^2)^2 ?

Answer 15

(43200000)/(36+t^2)^4 ?

Answer 16

incorrect

Answer 17

Let

u = 43,200,000t
v = (36+t^2)^2


so,

du/dt = ??
dv/dt = ??

Answer 18

du/dt = 43200000
dv/dt = I'm not sure about this one

Answer 19

use the chain rule to find dv/dt

Answer 20

I git 144t + 3t^2

Answer 21

v = (36+t^2)^2

dv/dt = 2(36+t^2)*(2t)

dv/dt = 4t(36+t^2)

dv/dt = 4t^3+144t

Answer 22

use du/dt and dv/dt to find the derivative of (43200000t)/(36+t^2)^2

Answer 23

I honestly don't understand

Answer 24

none of the problem except the first part

Answer 25

my Professor never went over this problem at all

Answer 26

so you've never gone over local min, local max and things like that?

Answer 27

not deep into like now we just when over basic problems, not word

Answer 28

The idea of this entire problem is this

We have some sales function x(t). The derivative, x ' (t), represents how fast sales occur instantaneously at any given time t. This derivative function is an upside down parabola. It peaks out at some point. This peak is the point we're interested in. This is the point where the speed peaks out and reaches its maximum.

How do we find this peak? We derive x ' (t) to get x '' (t) and set that equal to 0. We want to find when the tangent has a horizontal line.

Answer 29

So we have x ' (t) = (43200000t)/(36+t^2)^2

derive it using the quotient rule

u = 43200000t
u' = 43200000

v = (36+t^2)^2
v' = 4t^3+144t

Answer 30

x'(t) = u/v

x''(t) = (u'*v - v'*u)/( v^2 )

x''(t) = [ 43200000*(36+t^2)^2 - (4t^3+144t)*(43200000t)]/( (36+t^2)^2)^2

x''(t) =-(129600000*(-12+t^2))/(36+t^2)^3

I skipped a bunch of steps, but this gives you the basic idea

Answer 31

Now we set x''(t) equal to zero and solve for t

-(129600000*(-12+t^2))/(36+t^2)^3 = 0

-129600000*(-12+t^2) = 0

t = ??? or t = ???

Answer 32

The computer gives me 2sqrt 3

Answer 33

same here

Answer 34

so at t = 2*sqrt(3) which is approximately t = 3.464101615, the sales speed is maxed out

Answer 35

to find what the speed is at t = 3.464101615, you plug t = 3.464101615 into x'(t) which is the derivative function

Answer 36

Thank you i will write all this down to keep it for future reference:) I have never seen a problem like this before..

Answer 37

to find the amount of sales done at that point in time, you plug t = 3.464101615 into x(t). You may have to round

Answer 38

Thank You!

Answer 39

you're welcome