A piece of cardboard is 1 meter by 1/21/2 meter. A square is to be cut from each corner and the sides folded up to make an open-top box. What are the dimensions of the box with maximum possible volume?

Answer:The maximum dimensions of the box:Length of the box  = 0.788676 mBreadth of the box  = 0.288676 mStep-by-step explanation:Original piece of cardboard is a square with sides of length s.Length of the card board = l = 1 mBreadth of cardboard = b = 1/2 m = 0.5 mSquares with sides of length x are cut out of each corner of a rectangular cardboard to form a box.Now, length of the box = L = 1 - 2xAnd breadth of the box = B = 0.5 - 2xHeight of the box ,H = xVolume of the box ,V= L × B × H[tex]\frac{dV}{dx}=\frac{(1 -2x)(0.5-2x)x}{dx}[/tex][tex]\frac{dV}{dx}=\frac{d(0.5x-3x^2+4x^3)}{dx}[/tex][tex]\frac{dV}{dx}=0.5-6x+12x^2[/tex][tex]\frac{dV}{dx}=O[/tex]x = 0.394338  , 0.105662[tex]\frac{d^2V}{(dx)^2}=-6+24x[/tex]When , [tex]x=0.105662[/tex] , [tex]\frac{d^2V}{(dx)^2}<0[/tex] (maxima)The maximum dimensions of the box:Length of the box = L = 1 - 2x = 1 - 2(0.105662) m = 0.788676 mBreadth of the box = B = 0.5 - 2x = 0.5 - 2(0.105662) m = 0.288676 m