Three large squares of tin, each with edges 1m long, have four small, equal squares cut from their corners. All twelve resulting small squares are to be of the same size. The three large cross-shaped pieces are then folded and welded to make boxes with no tops, and the twelve small squares are used to make two small cubes. How should this be done to maximize the total volume of all five boxes?

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My working:

\(\displaystyle V=2x^3+3(1-2x)^2x\)

\(\displaystyle V=2x^3+3(1+4x^2-4x)x\)

\(\displaystyle V=2x^3+3x+12x^3-12x^2\)

\(\displaystyle V=14x^3-12x^2+3x\)

\(\displaystyle \frac{dV}{dx}=0\) for maxima.

\(\displaystyle 42x^2-24x+3=0\)

\(\displaystyle 14x^2-8x+1=0\)

\(\displaystyle x=0.3867\) or \(\displaystyle x=0.1846\)

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However, the answer given in the book is:

\(\displaystyle 0.25 m^3\) (all cubes, no open topped boxes)

Please help!

------------------------------------------------------------

My working:

\(\displaystyle V=2x^3+3(1-2x)^2x\)

\(\displaystyle V=2x^3+3(1+4x^2-4x)x\)

\(\displaystyle V=2x^3+3x+12x^3-12x^2\)

\(\displaystyle V=14x^3-12x^2+3x\)

\(\displaystyle \frac{dV}{dx}=0\) for maxima.

\(\displaystyle 42x^2-24x+3=0\)

\(\displaystyle 14x^2-8x+1=0\)

\(\displaystyle x=0.3867\) or \(\displaystyle x=0.1846\)

------------------------------------------------------------

However, the answer given in the book is:

\(\displaystyle 0.25 m^3\) (all cubes, no open topped boxes)

Please help!

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