Assume that a can is a perfect cylinder

The surface area of a cylinder with radius `r` and height `h` is given by

`S(r,h) = 2(pir^2) + h(2pir)`

this is twice times the area of the circle at the bottom of the can (to give surface area of the top...

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Assume that a can is a perfect cylinder

The surface area of a cylinder with radius `r` and height `h` is given by

`S(r,h) = 2(pir^2) + h(2pir)`

this is twice times the area of the circle at the bottom of the can (to give surface area of the top and the bottom) and the circumference of the can times the height (the sides are made up of a stack of `h` circles with radius `r` and height one unit).

The volume of the cylinder is given by

`V = h(pir^2)`

which is the area of a stack of `h` circles with radius `r` and height one unit.

We know the volume of the new cans - 320ml, so we have

`V = 320`

`implies` `h(pir^2) = 320`

Now we want to minimize the surface area with respect to `r` and `h`.

Use the method of Lagrange multipliers, ie simultaneously solve

1) ` (dS)/(dr) + lambda(dV)/(dr) = 0`

2) ` (dS)/(dh) + lambda(dV)/(dh) = 0`

subject to the constraint

`h(pir^2) = 320`

We have

1) `4pir + 2pih + 2pirhlambda = 0`

2) `2pir + pir^2lambda = 0`

Now 2) `implies` `r = -2/lambda`

Plugging this into 1) gives `h = -4/lambda`

So we need to satisfy the constraint

`-4/lambda(4pi/lambda^2) = 320`

`implies` `lambda^3 = -(16pi)/320 = -pi/20` `implies` `lambda = -root(3)((pi/20))` (ml)^(1/3) = (cm^3)^(1/3) = cm

(a millilitre is a cubic centimetre)

`implies` ` `` r=2root(3)((20/pi)) = 3.71` cm **answer**

`implies` `h = 4root(3)(20/pi) = 7.41` cm