Respuesta :
Answer:
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 ,
[tex]V=l^2*h[/tex]
[tex]=(40-2x)^2*x[/tex]
[tex]V=(40^2-2*2x*40+4x^2)*x[/tex]
[tex]V=(1600-160x+4x^2)*x[/tex]
[tex]V=4x^3-160x^2+1600x[/tex]
To get Maximum volume differentiate w.r.t 'x'
[tex]V'=12x^2-320x+1600[/tex]
[tex]V"=24x-320[/tex]
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
[tex]V=(40-2x)^2*x[/tex]
[tex]V=[40-2(20/3)]^2*20/3[/tex]
[tex]V=[40-40/3]^2*6.667[/tex]
=[tex][26.667]^2*6.667[/tex]
=4740.95
=[tex]4741 cu inches.[/tex]
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.
