Question

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.

Answer 1

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

Answer 2

Answer:

The dimensions are $35\times 14\times 5$ and the volume is 2450 inches³.

Step-by-step explanation:

Given : 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.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '45-2x'.

The breadth of the box is '24-2x'.

The volume of the box is $V=L\times B\times H$

$V=(45-2x)\times (24-2x)\times x$

$V(x)=4x^3-138x^2+108x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-138(2x)+108$

$V'(x)=12x^2-276x+108$

$V'(x)=12(x^2-23x+90)$

The critical point when V'(x)=0

$12(x^2-23x+90)=0$

$x^2-23x+90=0$

$(x-18)(x-5)=0$

$x=18,5$

18 is not possible we reject.

So, the height is 5 inches.

Derivate again w.r.t x,

$V''(x)=24x-276$

$V''(5)=24(5)-276$

$V''(5)=120-276$

$V''(5)=-156$

i.e. V(x) is maximum at x=5.

The dimensions are

Height = 5 inches

Length = 45-2(5)=35 inches

Breadth = 24-2(5)=14 inches.

The maximum volume is

$V=35\times 14\times 5$

$V=2450$

So, The dimensions are $35\times 14\times 5$ and the volume is 2450 inches³.