Question

An open-top rectangular box is constructed from a 10-in.- by-16-in. piece of cardboard by cutting squares of equal side length from the corners and folding up the sides. Find analytically the dimensions of the box of largest volume and the maximum volume?

Answer 1

The base of the box will have a length of `12` inches, a width of `6` inches, a height of `2` inches and a maximum volume of `144` cubic inches.

Explanation:

Let length of box =`x`, width of the box = `y` and the height of the box = `a` . [Note that the height of the box is also equal to length of the square cut out at the corners of the flat sheet from which it is constructed.]

So volume =`xya` . Considering the length of the box, it is seen that `[[16-x]]/2=a..........`[1] and that #[[10 - y]]/2`=a..........`[2]#

Equating `[1]` and `[2] ,` `[[16-x]]/2`= `[[10-y]]/2`, and solving for `y` yields `y=[x-6]`........`[3]`

So volume of box `V` =`x[x-6]a` , but from .....`[1]` `a=[[16-x]]/2`

Thus volume of box in terms of `x`, `V =x[x-6][[16-x]]/2`,

=`[x^2-6x][8-x/2]` = [`8x^2- x^3/2-48x+6x^2/2` `........`[4]#

Differentiating .....`[4]` wrt`x`, `[dV]/dx = 16x-3x^2/2-48+12x/2` = `0` for max/ min . Solving this expression, `[x-12][x-8/3]= 0` ,ie, `x= 12, x= 8/3`. Noting that the second derivative of `A` wrt`x` = `22-3x`, when `x= 12`, the second derivative is negative[ =`-14`] which indicates that this value of `x` ill maximise the volume of the box.

When `x= 12`, [from `[1]]` `a=[16-12]/2` = `2`. And from ...`[2]`, `[10-y]/2= a`, which when evaluated for `a=2` gives a value of `y` =`6`.

`V=[2][6][12]=144.` Hope this was helpful.