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.
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.