Question

A piece of cardboard is 27 inches by 54 inches. a square is to be cut from each corner and the sides folded up to make an open-top box. what is the maximum possible volume of the box? round your answer to the nearest four decimal places.

Answer 1

Let's call the side of the square $x$

When we cut out the square from each corner, the base of the box is

$(27 - 2x) \textrm{ by }(54 - 2x)$

and the height of the box is of course $x$ so the volume is

$V(x) = x(27-2x)(54-2x) = 4 x^3 - 162x^2 + 1458x$

$0 = V'(x) = 12x^2 - 324 x + 1458 = 6 (2 x^2 - 54 x + 243)$

$x = \frac 1 {2} (27 \pm \sqrt{27^2 - (2)(243)}) = \frac 1 2(27 \pm 9\sqrt{3})$

Approximately, x≈5.70577 or x≈21.294

x can't be bigger than half of 27, so we reject the second value.

We check the second derivative; a negative second derivative indicates a local maximum.

$V''(x) = 24x - 324$

$V''(5.70577 ) = 24(5.70577 ) - 324= -187.06$

Good. For our answer we want

$V(\frac 1 2(27 - 9\sqrt{3}) ) \approx 5.70577 (27 - 2(5.70577 ))(54-2(5.70577)) = 3787.99511615313...$

Answer: 3787.9951 cubic inches