# Determine the dimensions of 320ml cans which have a minimum surface area (full question contained)To avoid increasing the cost of 375ml cans of soft drink, the manufacturers have decided to reduce...

**Determine the dimensions of 320ml cans which have a minimum surface area** (full question contained)

To avoid increasing the cost of 375ml cans of soft drink, the manufacturers have decided to reduce the volume of new cans to 320ml.

Also, they have decided to keep to a minimum the surface area of the can to maximise profits.

Determine the dimensions of these new 320ml cans which have a minimum surface area.

Even if you just explain how to do it, it would be a MASSIVE help. Thank you!!!

*print*Print*list*Cite

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