Let the dimensions of the poster are L and W.

==> Then the area of the poster is :

A = L*W = 240

Now the dimensions with margins are:

Bottom and top : (w-2)

sides : (L-3)

==> Now the area is:

A= (w-2)(L-3)

But `L= 240/W`

`==gt A =...

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Let the dimensions of the poster are L and W.

==> Then the area of the poster is :

A = L*W = 240

Now the dimensions with margins are:

Bottom and top : (w-2)

sides : (L-3)

==> Now the area is:

A= (w-2)(L-3)

But `L= 240/W`

`==gt A = (w-2)(240/w - 3) `

`==gt A = (24 - 3w - 480/w + 6) `

`==gt A= -3w - 480/w + 30`

Now we will find the maximum value.

`==gt A'= -3 + 480/w^2 = 0 `

`==gt 480/w^2 = 3 ==gt w^2 = 480/3 = 160 `

`==gt w = sqrt160 = 4sqrt10 `

`==gt L= 480/w = 480/(4sqrt10) = 120/sqrt10= 12sqrt10`

Then, the dimensions that gives the maximum area are:

`4sqrt10 ` and `12sqrt10` .