Question

A large cylindrical soup has a volume of 3456pie cm^2. Find the radius and the height of the can that requires a minimum amount of material to construct it V=pie r^2 h and S= 2 pie r^2 + 2pie r h. OPTIMIZATION PROBLEM

Answer 1

pleaseeeeee.........

Answer 2

so you want the smallest surface area

Answer 3

yeah

Answer 4

how do i go about it

Answer 5

\[ S= 2 \pi r^2 + 2\pi r h=2\pi r(r+h)\]

Answer 6

yeah

Answer 7

\[V=\pi r^2\]

Answer 8

V=pie r^2h

Answer 9

whops
\[V=\pi r^2h\]

Answer 10

yeah

Answer 11

\[V=3456\pi [\text{cm}]^2\]
so can you find \[r^2h=\]

Answer 12

then find \(h(r)=\)

Answer 13

How?

Answer 14

did you find r^2h?

Answer 15

r^2h=3456

Answer 16

now divide by r^2

Answer 17

what divide by r^2

Answer 18

\[r^2h=3456\]
\[h=\frac{3456}{r^2}\]\[h(r)=\frac{3456}{r^2}\]

Answer 19

yeah..

Answer 20

\[S=2\pi r\left(r+\frac{3456}{r^2}\right)\]

Answer 21

sure am gettin it

Answer 22

now differentiate \(S\) with respect to \(r\)

Answer 23

if might be easier to differentiate if you expand the brakets first

Answer 24

i.e 2pie r^2 +6912pie r/r

Answer 25

\[S=2\pi \left(r^2+\frac{3456}{r}\right)\]
\[\frac{\text dS}{\text dr}=\]

Answer 26

how should i find the derivate

Answer 27

\[y=x^n\]\[\frac{\text dy}{\text dx}=nx^{n-1}\]

Answer 28

wats the derivative of 2pie?

Answer 29

\[2\pi=2\pi\times1 =2\pi x^0\]

Answer 30

are you saying is constant i tot is -0

Answer 31

the derivative of a constant is zero yes

Answer 32

so the derivative is 2r+3456r^-2

Answer 33

the derivative is
\[\frac{\text dS}{\text dx}=2\pi\left(2r+\frac{3456}{r^2}\right)\]

Answer 34

you said the derivative to 2pir is =0

Answer 35

dont forget the chain rule
\[(fg)'=f'g+gf'\]

when the derivative is zero there is a critical point in the solution curve

Answer 36

(i mean the product rule not the chain rule )



now. when is
\[\left(2r+\frac{3456}{r^2}\right)=0\]
solve for r

Answer 37

r=-12

Answer 38

yeah, how'd you get that ?
your right by the way)

Answer 39

except the radius cant really be negative

Answer 40

i solved for r as you said. yeah plus or negative r. there fore r=12

Answer 41

ops i made a mistake on the derivative it should be \[\frac{\text dS}{\text dx}=2\pi\left(2r-\frac{3456}{r^2}\right)\]so the critical point is when\[2r-\frac{3456}{r^2}=0\]\[2r=\frac{3456}{r^2}\]\[r^3=\frac{3456}{2}\]\[r=\sqrt[3]{\frac{3456}{2}}=\]

Answer 42

ok r=12 how we solve fot h now.

Answer 43

yes!

Answer 44

how should i solve for h

Answer 45

A Mathematica solution is attached.

Answer 46

\[r=12\]

\[h(r)=\frac{3456}{r^2}\]
\[h(12)=\frac{3456}{12^2}\]
\[h=\frac{3456}{144}=\dots\]

Answer 47

h=24. thanks alot and to robotobey thanks alot too.

Answer 48

the general case of optimisation of a cylinder is
h=2r