Question

Some help with calculus optimization please!!

Answer 1

I have done problems 1-7.

Answer 2

@agent0smith

Answer 3

I can do number 8 on my own

Answer 4

Can someone help with number 9?

Answer 5

Okay the formula is \[A(t)=Ce ^{k \sqrt{t}+r(40-t)}\]

Answer 6

and the domain is 0≤t≤41

Answer 7

Just find the derivative, same as last time. It's not much different, save for r and k.

Answer 8

yeah my bad typo

Answer 9

\[A'(t)=Ce ^{k \sqrt{t}+r(40-t)}(\frac{ k }{ 2\sqrt{t} }-r)\]

Answer 10

Do i find the maximum value now?

Answer 11

Yep, same way as last time.

Answer 12

Set it equal to zero right?

Answer 13

Yep

Answer 14

\[t=\frac{ k }{ \sqrt{2}*r }\]

Answer 15

is that right?

Answer 16

Doesn't look right... solve \[\large \frac{ k }{ 2\sqrt{t} }-r=0\]

Answer 17

\[r=\frac{ k }{ \sqrt{2}t }\]

Answer 18

whoops

Answer 19

√t*2

Answer 20

You're solving for t.

Answer 21

okay

Answer 22

Idk how you forgot that lol. It's the same steps as last time.

Answer 23

\[t=\frac{ k ^{2} }{ 4r ^{2} }\]

Answer 24

is this right @agent0smith

Answer 25

Looks right

Answer 26

okay cool so is this topt?

Answer 27

Yep

Answer 28

Okay cool thanks mate!!

Answer 29

@agent0smith for # 10 do i just put in random values or do you think there is certain values that they might be looking for, to prove that my topt equation is right

Answer 30

Random values, and you could check your graphs maximum lines up with the t value you found above (you already know it will)

Answer 31

okay thanks