Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 29 in. by 16 in. by cutting congruent squares from the corners and folding up the sides. then find the volume.

height = 3.3 in

V = 694.848 in^3

Step-by-step explanation:

length of cardboard, l = 29 in

width of cardboard, w = 16 in

let the side of square is d.

so the length of box = 29 - 2y

width of box = 16 - 2y

height of box = y

Volume of box = length x width x height

V = (29 - 2y) x (16 - 2y) x y

V = 464 y - 90 y^2 + 4 y^3

dV/ dy = 464 - 180 y + 12y^2

For maxima and minima, dV/dy = 0

12y^2 - 180 y + 464 = 0  

By solving

y = 3.3 in, 11.7 in

Now find double differentiation

d²V/dy² = 24 y - 180

Put, y = 3.3 in, it is - 100.8

Put, y = 11.7 in, it is + 100.8

So for maximum volume, y = 3.3 in

And the value of maximum volume

V = 464 (3.3) - 90(3.3 x 3.3) + 4 (3.3 x 3.3 x 3.3) = 1531.2 - 980.1 + 143.748

V = 694.848 in^3

robert puts 2 in each row

no problobm

step-by-step explanation: