Question

A cylindrical can, open at the top, is to hold 590cm^3 of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can.

Answer 1

pi*r^2*h = 590
hence h = 590/(pi*r^2)
Now that we know what r is
what we need is the surface area.
that is 2*pi*r*h + pi*r^2
replace h with r
2*590/r + pi * r^2 = 1180/r + pi*r^2
Now for minimum value, differentiate and solve for 0
hence
2pi*r - 1180/r^2 = 0
when you solve this, you get 5.73 which is the radius. and the height is 5.73 as well...

Answer 2

how did you get the height

Answer 3

where is the second derv? or do we need it in this case?

Answer 4

i plugged the value of r which i got by differentiating the equation into the equation h = 590/(pi*r^2)
which is
590/(pi*(5.73^2))
Second derivative won't be necessary. To find the maximum value you just take first derivative for minimum value :)

Answer 5

you basically do
dSA/dr right. Hence you will get minimum/maximum value in the first derivative itself.

Answer 6

thank you!!!! but wait what would be the equation to solve for h

Answer 7

i mean r

Answer 8

Which grade are you in? And have you learnt differentiation? I used differentiation for this problem so i better know this beforehand.

Answer 9

i know what differentiation is

Answer 10

just what is r=
and h=
exactly not rounded is what i mean

Answer 11

Oh then honestly this will be hard to explain. What i basically did was take the gradient of the function i derived, which was 1180/r + pi*r^2
using differentiation. how that we had the gradient, equating it to 0 would naturally give us the lowest possible value for the variable.
How i differentiated was
d/dx(1180*x^(-1) + pi*r^2 ) = -1180*x^(-2) + 2pi*r
now when i solve this for 0 i got r = 5.73
Sorry but this is really hard to do without knowledge of differentiation. :/

Answer 12

no rounding the answer

Answer 13

I can help you through this and explain it if you want to understand.

Answer 14

yesss!!! please

Answer 15

"A cylindrical can, open at the top, is to hold 590cm^3 of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can."

So using this information what kind of formulas can we make? Well it looks like the volume and surface area are important. Do you know how to make these formulas?

Remember, V=l*w*h and A=l*w. Just add up all the areas of the faces. The volume should be slightly easier. Show me what you get and if you have trouble coming up with these I'll help you.

Answer 16

well its a cylinder so v=pi*r^2*h & SA=2πrh+2πr^2

Answer 17

did i do it right??

Answer 18

Yeah, perfect. So what do you know about this problem? Well since you know Volume you can plug that in to your equation.

From there you have two equations, both have R and H, so you can solve for Area in terms of just one part of it, the radius or the height. So if you plug one equation into the other you basically have Area as a function that can change depending on what you pick for your height/radius.

So go ahead and plug in your volume equation to your area equation and show me what you get. Pick whichever you think will be easiest to take the derivative of and I'll explain why in a second.

Answer 19

do i take the derv first?

Answer 20

if not is is sqroot(pi*h)^-1=h
[im confused]

Answer 21

?

Answer 22

Nope don't take the derivative first, instead you need Area as a function of one variable, your height or radius.

so, A(r) or A(h)

See, since you have the volume equation, if you plug in the radius you'll get the height. So they're connected now.

Basically you're making a function of Area that will be like: