Question

The region R is a rectangle with vertices P (a,ln a), Q (a, 0), S (3, 0), and T (3, ln a), where 1 < a < 3.

Answer 1

B. The area of the rectangle is maximized for some c between 1 and 3. Write the expression you would need to solve in order to find c. You don’t need to find c.

C. What is the rate of change of the area when a =3, if a is decreasing at a rate of da/dt = − 0.5 units / sec? You don’t need to simplify, of course. Use appropriate units.

D. Again, a is decreasing at the rate of da/dt = − 0.5 units / sec. At a = 2.9, is the area of the rectangle increasing or decreasing? How do you know? Use appropriate units in your answer.

Answer 2

Confirm my answers please:

C. A = (3 - a)•ln(a)
dA/dt = -ln(a)(da/dt) + (3 - a)(da/dt)/a
= -ln(3)(-.5)un/sec + (3 - 3)(-.5)un/(3sec)
= .5•ln(3)un/sec
= ln(√3) units/sec

D. dA/dt = -ln(a)(da/dt) + (3 - a)(da/dt)/a
= -ln(2.9)(-.5)un/sec + (3 - 2.9)(-.5)un/(2.9sec)
= .5•ln(2.9)un/sec - .05un/(2.9sec)
aprox. = 0.515units/sec
So the area of the rectangle is increasing a = 2.9.

Answer 3

I am not sure how to do B.

Answer 4

hey so why do you think A is increasing as a decreases?

Answer 5

Because 0.515 units/sec is positive, so it increases? Right?

Answer 6

@myininaya

Answer 7

ok i was missing a negative
but yeah it looks good

Answer 8

but is .515 correct?

Answer 9

Well.. that's why I am here to confirm that lol.

Answer 10

yep your arithmetic is right

Answer 11

Can you also help me with B? Not sure how to approach that.

Answer 12

sounds like you need to use the mean value theorem

Answer 13

We learned that during first semester. I don't recall how to do the mean value theorem.

Answer 14

Let A be continuous function on [1,3] and differentiable on (1,3) so that there is a c in (1,3) such that \[f'(c)=\frac{f(3)-f(1)}{3-1}\]

Answer 15

where f is A

Answer 16

So we just plug in 3 and 1 into the equations and solve algebraically and we have the answer?

Answer 17

and plug in c into the derivative for A

Answer 18

Okay, I know it says that we don't need to find C but let me just quickly find the answer and confirm with you.

Answer 19

well we don't know that derivative for a is constantly -.5 do we ?

Answer 20

Oh. Lol, right. Nvm then. I think that's it ^_^ Thanks for your help.

Answer 21

well you could try to
you actually can reduce the equation more

Answer 22

try it

Answer 23

but I don't think we can solve it using our algebra skills
we need a calculator
or i guess we could use a calculus way to approximate the value for c

Answer 24

go in assuming you don't know the rate of a at c

Answer 25

Okay. I'll attempt it lol. One sec.

Answer 26

I feel dumb for asking this but what are the equations for f(3) and f(1)? :0

Answer 27

Well I said A is f
and it still is remember A(a)=(3-a)ln(a)

Answer 28

Oh I see, so we just plug in 3 and 1 as a?

Answer 29

yep

Answer 30

(3-3)/Ln(3) - (3-1)/Ln(1) = 0/Ln(2) - 2/Ln(1) ?

Answer 31

you should see we are working with A'(c)=0
since 3-3=0 and ln(1)=0

Answer 32

Oh wait, opps. Mistake. Hold on.

Answer 33

\[A'(c)=\frac{A(3)-A(1)}{3-1} \]

Answer 34

I meant to put -2Ln(1) / 2, idk why I put the division symbol lol -_-

Answer 35

since A(3) = 0, right?

Answer 36

\[A'(c)=\frac{(3-3)\ln(3)-(3-1)\ln(1)}{3-1}\]

Answer 37

A'(c)=0 since ln(1)=0 and 3-3=0

Answer 38

Fair enough. Final answer (if it had asked for it) is 0, right? :)

Answer 39

nope

Answer 40

you have to solve A'(c)=0 for c
but told you not too
What is A'(a)= you actually already gave this way above

Answer 41

Ah, I see. Well thanks for your help anyways.

Answer 42

replace the a with c

Answer 43

\[A'(c)=\frac{da}{dt}|_{a=c}(\frac{3-c}{c}-\ln(c))\]
right?

Answer 44

And that equals zero

Answer 45

so when you have a*b=0 this means a=0 or b=0

Answer 46

Hm, that's very informative. Again thanks for your help. But if we didn't actually try to solve it, the actual expression needed to solve would be A'(c) = A(3) - A(1)/(3-1) and I can just leave it at that?

Answer 47

well yeah
i don't know how simplified they want it

Answer 48

I think that's good enough because the question doesn't go into specifics.

Answer 49

alight you could say A'(c)=0 at least
it is much shorter

Answer 50

Ok.