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Let x cm = side of square to be removed

Volume of box (v) will be x(10 - 2x)(10 - 2x)

v = 4x3 - 40x2 + 100x

v changes in accordance with f(v), the derivative of this, which is 12x2 - 80x + 100

When the rate of change is zero then the volume has reached a maximum or minimum ie when f(v) equals zero.

F(v) factorises as (6x - 10)(2x - 10) = 0

ie the values of x are 10/6 or 5, the latter being meaningless as there would be no box!

The squares to be removed are thus 1 and two thirds cm a side, giving the box dimensions of 6 and two thirds square by 1 and two thirds cm deep, a total volume

of 74 and two twentysevenths cm3 (74.074074 cm3)

Check: if square is 1.666 cm, vol is 6.668 x 6.668 x 1.666 = 74.074065

If square is 1.668 cm then vol is 6.664 x 6.664 x 1.668 = 74.074038

The "maximum volume" value thus lies between 1.666 and 1.668, confirming our calculation of 1.667 cm.

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Q: What is the maximum volume of a box you can make by cutting out squares from each corner of a 10 cm by 10 cm sheet?
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