Question

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 39 in. by 21 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.

Answer 1

Length=29.8 inches

Width=11.8 inches

Height=4.6 inches

Volume=1,617.54 cubic inches

Explanation:

Let the side of congruent square cut =x inches

So the length of the rectangular box=(39-2x)

width = (21-2x)

height = x

The volume V=Length*Width*Height

= (39-2x)*(21-2x)*x

dV/dx= (39-2x)(21-4x)-2x(17-2x)=0

Simplify the equation above

819-156x-42x+8x^2-34x+4x^2=0

We have,

12x^2 -232 +819=0

Solve the quadratic equation using formula

a=12

b= -232

c=819

x= -b +or- √b^2-4ac/2a

= -(-232) +- √(-232)^2 - (4)(12)(819) / (2)(12)

= 232 +or- √53824 - 39312 / 24

= 232 +or- √14512 / 24

= 232 +or- 4√907 / 24

x= 232 / 24 + 4√907 / 24

=14.6861

Or

x=232 / 24 - 4√907 / 24

=4.64726

x=4.6 inches

Length=(39-2x)

={39-2(4.6)}

= 29.8 inches

Width=(21-2x)

={21-2(4.6)}

= 11.8 inches

Height=x= 4.6 inches

Volume=(39-2x)*(21-2x)*x

={39-2(4.6)}*{21-2(4.6)*4.6

=(39-9.2)*(21-9.2)*4.6

=29.8*11.8*4.6

=1,617.544

Approximately 1,617.54

Volume=1,617.54 cubic inches