Answer:
Hence the maximum possible volume will be the 778.53 c.c
Step-by-step explanation:
Given:
A rectangle with 13 x 26 dimensions
And corners are cut to form side squares.
To Find:
Maximum possible volume for box
Solution :
Consider a rectangle of 13 x 26 dimension with and side of square at corner be x.
(Refer the attachment)
Now,
Formulating the volume equation for the box
So corner square sides we are going to fold up which makes height of the box
and remaining part will be length and breadth
As shown in fig,
Length=26-x
breadth=13-x
And height will be x
[tex]V(x)=x*(26-x)*(13-x)[/tex]
To get maximum volume differentiate the above equation,
[tex]V(x)=x*(26*13-26*x-13*x+x^2)[/tex]
[tex]V(x)=x^3-39x^2+338x\\[/tex]
[tex]V'(x)=3x^2-78x+338[/tex]
[tex]V''(x)=6x-78[/tex]
Now ,Solve the Quadratic Equation to get x values,
[tex]3x^2-78x+338[/tex]=0
x=[-b±(b^2-4ac)^1/2]/2a
x=[78±Sqrt[(78)^2-4*338*3)]/2*3
x=[78±Sqrt(3028)]/6
x=[78±55.027]/6
x=78+55.027/6 or x=78-55.027/6
x=22.17 or x=3.8288
Use these values in 6x-78 to know which value posses the max and min value for the function.
So when x=22.17
6x-78=6*22.17-78
=55.02>0 i.e function will have minimum value .
When x=3.8288
6*3.8288-78
=-55.0272<0 i.e. Function will have maximum value
Now, the function will defines the maximum volume
[tex]V(x)=x^3-39x^2+338x[/tex]
[tex]V(x)=3.8288^3-39*(3.82883)^2+338*3.8288[/tex]
[tex]V(x)=56.13-571.73+1294.13[/tex]
[tex]V(x)=778.53 C.C[/tex]
The volume of a box is the amount of space in it.
The maximum value of the box is [tex]\mathbf{422.8215in^3}[/tex]
The dimension of the cardboard is given as:
[tex]\mathbf{Length = 13}[/tex]
[tex]\mathbf{Width = 26}[/tex]
Assume the cut-out is x.
So, the dimension of the box is:
[tex]\mathbf{Length = 13 -2x}[/tex]
[tex]\mathbf{Width = 26 - 2x}[/tex]
[tex]\mathbf{Height = x}[/tex]
The volume of the box is:
[tex]\mathbf{V = (13 -2x) (26 - 2x)x}[/tex]
Expand
[tex]\mathbf{V = 338x - 52x^2 - 26x^2 + 4x^3}[/tex]
[tex]\mathbf{V = 338x -78x^2 + 4x^3}[/tex]
Differentiate
[tex]\mathbf{V' = 338 -156x + 12x^2}[/tex]
Set to 0
[tex]\mathbf{338 -156x + 12x^2 = 0}[/tex]
Rewrite as:
[tex]\mathbf{12x^2 -156x + 338 = 0}[/tex]
Divide through by 2
[tex]\mathbf{6x^2 -78x + 169 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = (2.75, 10.25)}[/tex]
10.25 is greater than the dimensions of the box.
So, the possible value of x is:
[tex]\mathbf{x = 2.75}[/tex]
Recall that:
[tex]\mathbf{V = 338x -78x^2 + 4x^3}[/tex]
So, we have:
[tex]\mathbf{V =338 \times 2.75 - 78 \times 2.75^2 + 4 \times 2.75^3}[/tex]
[tex]\mathbf{V =422.8215}[/tex]
Hence, the maximum value of the box is [tex]\mathbf{422.8215in^3}[/tex]
Read more about volumes at:
https://brainly.com/question/13529955
