Respuesta :
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
Answer:
The dimensions are [tex]35\times 14\times 5[/tex] 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 [tex]V=L\times B\times H[/tex]
[tex]V=(45-2x)\times (24-2x)\times x[/tex]
[tex]V(x)=4x^3-138x^2+108x[/tex]
Derivate w.r.t x,
[tex]V'(x)=4(3x^2)-138(2x)+108[/tex]
[tex]V'(x)=12x^2-276x+108[/tex]
[tex]V'(x)=12(x^2-23x+90)[/tex]
The critical point when V'(x)=0
[tex]12(x^2-23x+90)=0[/tex]
[tex]x^2-23x+90=0[/tex]
[tex](x-18)(x-5)=0[/tex]
[tex]x=18,5[/tex]
18 is not possible we reject.
So, the height is 5 inches.
Derivate again w.r.t x,
[tex]V''(x)=24x-276[/tex]
[tex]V''(5)=24(5)-276[/tex]
[tex]V''(5)=120-276[/tex]
[tex]V''(5)=-156<0[/tex]
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
[tex]V=35\times 14\times 5[/tex]
[tex]V=2450[/tex]
So, The dimensions are [tex]35\times 14\times 5[/tex] and the volume is 2450 inches³.
