Question Help Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 45 in. by 24 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.
Accepted Solution
A:
You need to cut 4 square corner pieces. Each corner square has side x. The length of the box will be 45 - 2x, and the width will be 24 - 2x. The height is x. The volume of the box is
(45 - 2x)(24 - 2x)(x) =
= (1080 - 90x - 48x + 4x^2)x
= 1080x - 138x^2 + 4x^3
V = 4x^3 - 138x^2 + 1080x
To find a maximum volume, we differentiate the volume equation above and set equal to zero. Then we solve for x.
dV/dx = 12x^2 - 276x + 1080
12x^2 - 276x + 1080 = 0
x^2 - 23x + 90 = 0
(x - 18)(x - 5) = 0
x = 18 or x = 5
x cannot be 18 because 24 - 2(18) is negative and the width of the box cannot be negative.
x = 5
The length of the box is: 45 - 2(5) = 45 - 10 = 35
The width of the box is: 24 - 2(5) = 24 - 10 = 14
The height is : 5
The volume is V = LWH = 35 in. * 14 in. * 5 in. = 2450 in.^3