Question

The rectangular beam is cut from a cylindrical log of radius 10 cm. The strength of a beam of width w and height h is proportional to wh^2 (see figure attached inside.) Find the width and height of the beam of maximum strength. Round the answers to 2 decimal places.

w=_______cm
h=_______cm

don't get where to start! :( @johnweldon1993 ?

Answer 1

figure!

Answer 2

Hmm...thinking on it lol

Answer 3

haha okay :P

Answer 4

Oh wait...I think I got it..

or at least I have some numbers haha...lets see if we can work it out!

Answer 5

okay! :)

Answer 6

Hmm so you have strength as a function of width and height,\[\Large\rm S(w,h)=w h^2\]And you want to maximize this function.

Hmmm

Answer 7

We need it in terms of one variable before we can try maximizing.
So you need to use the uhhhh.. radius information I think.. hmm

Answer 8

hahaa hmm :P

yeah this problem is so confusing :/

Answer 9

This should allow you to write w in place of h,
or the other way around.

Understand how I made that triangle? :o

Answer 10

yes i think so.. has to do with pytahgorean theorem right?

Answer 11

Yah I looked at the picture, the 10 radius, I extended it in both directions.
That gives us a nice triangle.

Yes, Pythagorean Theorem will let you substitute variables.

Answer 12

Ahh exactly...we can just do

\[\large W^2 + H^2 = 20^2\]

Answer 13

So just isolate one of the variables....

Answer 14

okay:)

Answer 15

Well actually...isolate h^2 (we'll want to sub that back into the

S = wh^2 equation

Answer 16

okay

so we get this?

h^2 = 20^2 - w^2

?

Answer 17

Right...so sub that into the S = wh^2 equation

\[\large S = w(20^2 - w^2)\]

Simplify that...

Answer 18

S=20w^2 - w^3

?

Answer 19

Well

\[\large S = 20^2w - w^3\]

*W would't be squared when multiplied to the 20^2*

Answer 20

Right?

Answer 21

ohh right my bad haha

Answer 22

Now just take the derivative of that function...

\[\large \frac{ds}{dw} = \space?\]

Answer 23

umm 400-3w^2 ?

Answer 24

Right ....so now we have to set that = 0

\[\large 400 - 3w^2 = 0\]

\[\large 3w^2 = 400\]

Think you can figure out what to do next :)

Answer 25

w^2 = 400/3
w=20/√3
?

Answer 26

Right "just remember they want a decimal answer...but yes keep this for now...

This is 'W'

What is 'H'

*Plug it into Pythagorea's

Answer 27

okay:)

h=20/√(2/3) ?

Answer 28

Well lets see..does

\[\large (\frac{20}{\sqrt{3}})^2 + (\frac{20}{\sqrt{\frac{2}{3}}})^2 = 400\]
?

\[\large \frac{400}{3} + \frac{400}{\frac{2}{3}} = 400\]

\[\large \frac{400}{3} + 600 = 400\]

Doesn't look right...

Lets try something else...

\[\large (\frac{20}{\sqrt{3}})^2 + h^2 = 400\]
\[\large \frac{400}{3} + h^2 = 400\]
\[\large 400 + 3h^2 = 1200\]
\[\large 3h^2 = 800\]
\[\large h^2 = \frac{800}{3}\]
\[\large h = \sqrt{\frac{800}{3}}\]
\[\large h = \sqrt{\frac{20^2 \times 2}{3}}\]
\[\large h = 20 \sqrt{\frac{2}{3}}\]

Answer 29

ohhh okay i see now :)

so now just round and there we have h and w? :O

Answer 30

Correct!

Answer 31

awesome!! thank you!! :D

Answer 32

No problem! :)