Question

AP Calculus:
A cylindrical can is to hold a volume of 400 cc. Find the dimensions (radius and height) of the can so the surface area of the whole can is a minimum

Answer 1

since V=pi*r^2*h
surface area A=2*pi*r*h + 2(pi*r^2)

solve that expression where d(A)/dr=0. That will give you a value for r where A is at a local maximum or minimum.

Now take the second derivative of A with repect to r. Plug in the value for r you found in solving for the root of d(A)/dr. Check the sign of the result. If d2(A)/dr2 at that value of r is positive, then what you found is a local minimum. That's what you want.

Answer 2

what is dA \[2\pi \times r \times h + 2\pi \times r^2\]

Answer 3

and what is dr? xD

Answer 4

@stgreen

Answer 5

A=2π×r×h+2π×r^2
differentiate it w.r.t r
dA/dr=2pi*h+4pi*r
put dA/dr=0 and find r....then find second derivative of A and follow as written in first reply

Answer 6

\[so \frac{ dA }{ dr } = 2\pi h + 4\pi r=0\]

Answer 7

@stgreen

Answer 8

yeah write r in terms of h.....and then find second derivative of A w.r.t r

Answer 9

so I take the second derivative of \[2\pi h +4\pi r\]
again? :)

Answer 10

@stgreen

Answer 11

again??lol you are taking its second derivative for first time....do it...i could do but mod kicks me out for telling answers instead of method

Answer 12

@stgreen

Answer 13

\[2\pi h + 4\pi r\]
sorry .. i meant that xD

Answer 14

that would be 2nd derivative?

Answer 15

noooooope it'll be 4pi which is always positive so you have a minima what you required

Answer 16

that would be 1st derivative of A? \[2\pi h + 4\pi r \]

Answer 17

then i take the derivative of that again :o

Answer 18

What you want to do is to minimize the expression you have for A with respect to the variable r or h. So use the expression you have that relates r to h (since the can is contrained to a fixed volume of 400) and substitute out the expression for A so it is an expression of only one variable r or h. It's probably easier to substitute out the h so A is an expression of r only.

Then take the derivative of A with respect r and solve that expression where d(A)/dr=0. That will give you a value for r where A is at a local maximum or minimum.

Now take the second derivative of A with repect to r. Plug in the value for r you found in solving for the root of d(A)/dr. Check the sign of the result. If d2(A)/dr2 at that value of r is positive, then what you found is a local minimum. That's what you want. That's the minimum in A. If d2(A)/dr2 at that value is negative, then what you found is actually a local maximum. Once you found the r that minimizes A, you can just use the formula to find the corresponding h. That gives you the can dimensions that minimizes material used while still holding volume of 400.

Answer 19

JUST FOLLOW IT

Answer 20

haha okay :P

Answer 21

what so funny??