MATH SOLVE

4 months ago

Q:
# 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

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