# a poster of area 6000cm^2 has a blank margins of width 10 cm on the top and bottom and 6cm on the sides. find the dimensions of the poster that maximize the printed area.

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### 1 Answer

You should come up with the notations for length and width: x=length and y=width.

Supposing that the shape of poster is rectangular, hence the area of the poster is `6000 = x*y` .

You need to evaluate the area of the printed zone such that:

`(x-12)(y-20)=A`

You may write the area of printed zone in terms of x such that:

`6000 = x*y =gt y = 6000/x`

`(x-12)(y-20)=A_(x,y) =gt A(x) = (x-12)(6000/x-20)`

Opening the brackets yields:

`A(x) = 6000 - 20x - 12*6000/x + 240`

`A(x) = 6240 - 20x - 72000/x`

You need to determine the maximized printed area, hence you need to differentiate the function of area with respect to x such that:

`A'(x) = -20 + 72000/x^2`

You need to remember that a function reaches its maximum at a point that represents the root of derivative of function, hence you need to solve for x equation A'(x)=0 such that:

`-20 + 72000/x^2 = 0 =gt -20x^2 + 72000 = 0 =gt `

`x^2 = 7200/2` `x^2 = 3600 =gt x = 60` cm

**Hence, the dimensions of poster that maximize the printed zone are of x = 60 cm and y = 100 cm.**