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:
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.