We have a rectangular sheet of paper; let the dimensions be `l,w` . The paper has 2cm margins at the top and bottom and 1cm margins on the sides, so the usable part of the paper is `(l-4)(w-2)` . We need to have 30 square centimeters of type.

We are asked to minimize the area `A` of the paper where `A=l*w` .

We know that `30=(l-4)(w-2) ==> w-2=30/(l-4) ==> w=(2l+22)/(l-4)`

So we want to minimize `A=l((2l+22)/(l-4))=(2l^2+22l)/(l-4)`

This is a rational function and so will be continuous and differentiable on its domain. This type of function will have any extrema at critical points, which are where the derivative of the function is zero or fails to exist. (Note that the domain is l>4 since the paper has nonnegative length.)

Taking the derivative of A with respect to l, using the quotient rule, we get:

`(dA)/(dl)=((l-4)(4l+22)-(2l^2+22l))/((l-4)^2)`

`=(4l^2+6l-88-2l^2-22l)/(l-4)^2`

`=(2l^2-16l-88)/((l-4)^2`

Now `(dA)/(dl)` is also rational and exists everywhere except l=4 which is not in the domain. So we set the derivative equal to zero and solve:

`2l^2-16l-88=0`

`2(l^2-8l-44)=0`

Completing the square we get:

`l^2-8l+16=44+16`

`(l-4)^2=60`

`l=4 pm sqrt(60)`

Since l>4 we get

`l=4+2\sqrt(15)`

Substituting we get `w=2+sqrt(15)/2`

Then the minimal area is

`A=(4+2sqrt(15))(2+sqrt(15))=38+8sqrt(15)~~68.98"cm"^2`

**We could use the first derivative test to confirm that the zero we found for the derivative is actually a minimum.**

See the attached file for a graph of the possible areas.

**Further Reading**