From a thin piece of cardboard 40 in. by 40 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? Round to the nearest tenth, if necessary.

The maximum dimension for box will be side length as 26.667 inches and height will be 6.667 inches and volume will be 4741 cu inches.

Step-by-step explanation:

Given:

With dimensions as

40x40 square and sides are folded in square corners.

To Find:

Maximum volume that will yield and maximum dimension for box

Solution:

Consider, a box with 40 x 40 dimensions as follows and

let a x side length square corners are cut and folded up and a new figure is formed .and for that what will be the maximum volume and dimension.

Now,

as the there 2 corner for each side i.e. 2 square on side length

Resultant length will be 40-2x

So volume is given by ,

To get Maximum volume differentiate w.r.t 'x'

Now solve for x with help of quadratic equation we get ,

12x^2-320x+1600

=4(3x-20)(x-20)

Therefore ,

3x-20=0 or x-20=0

x=20/3 or x=20

x=16.666 or x=20

Now use this values in a 2nd derivative equation,

V"=24(6.667)-320)

V"=160-320

V"=-160 <0 ,................equation(1)[Here maximum volume will be presented]

For X=20

V"=24*20-320

V"=480-320

=160>0 ..................Equation(2)[here minimum volume will presented]

Comparing equation 1 and 2 we get ,

Take as x=20/3 as the folded square length as it is gives minimum value for 2nd derivative.

So

=

=4740.95

=

So side lengths =40-2x=40-2*20/3

=40(1-1/3)

=40(2/3)

=26.667 inches

And the height will be x=20/3=6.667 inches.

what is the answer?