Let's call the side of the square [tex]x[/tex]
When we cut out the square from each corner, the base of the box is
[tex](27 - 2x) \textrm{ by }(54 - 2x)[/tex]
and the height of the box is of course [tex]x[/tex] so the volume is
[tex]V(x) = x(27-2x)(54-2x) = 4 x^3 - 162x^2 + 1458x[/tex]
[tex]0 = V'(x) = 12x^2 - 324 x + 1458 = 6 (2 x^2 - 54 x + 243)[/tex]
[tex] x = \frac 1 {2} (27 \pm \sqrt{27^2 - (2)(243)}) = \frac 1 2(27 \pm 9\sqrt{3}) [/tex]
Approximately, x≈5.70577 or x≈21.294
x can't be bigger than half of 27, so we reject the second value.
We check the second derivative; a negative second derivative indicates a local maximum.
[tex]V''(x) = 24x - 324 [/tex]
[tex] V''(5.70577 ) = 24(5.70577 ) - 324= -187.06[/tex]
Good. For our answer we want
[tex] V(\frac 1 2(27 - 9\sqrt{3}) ) \approx 5.70577 (27 - 2(5.70577 ))(54-2(5.70577)) = 3787.99511615313...[/tex]
Answer: 3787.9951 cubic inches
The volume of a box is the amount of space in it.
The maximum value of the box is [tex]\mathbf{ 3787.9934in^3}[/tex]
The dimension of the cardboard is given as:
[tex]\mathbf{Length = 27}[/tex]
[tex]\mathbf{Width = 54}[/tex]
Assume the cut-out is x.
So, the dimension of the box is:
[tex]\mathbf{Length = 27 -2x}[/tex]
[tex]\mathbf{Width = 54 - 2x}[/tex]
[tex]\mathbf{Height = x}[/tex]
The volume of the box is:
[tex]\mathbf{V = (27 -2x) (54 - 2x)x}[/tex]
Expand
[tex]\mathbf{V = 1458x- 162x^2 + 4x^3}[/tex]
Differentiate
[tex]\mathbf{V' = 1458- 324x + 12x^2}[/tex]
Set to 0
[tex]\mathbf{ 1458- 324x + 12x^2 = 0}[/tex]
Rewrite as:
[tex]\mathbf{ 12x^2- 324x +1458 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = (5.71, 21,29)}[/tex]
21.29 is greater than the dimensions of the box.
So, the possible value of x is:
[tex]\mathbf{x = 5.71}[/tex]
Recall that:
[tex]\mathbf{V = 1458x- 162x^2 + 4x^3}[/tex]
So, we have:
[tex]\mathbf{V = 1458(5.71) - 162(5.71)^2 + 4(5.71)^3}[/tex]
[tex]\mathbf{V = 3787.9934}[/tex]
Hence, the maximum value of the box is [tex]\mathbf{ 3787.9934in^3}[/tex]
Read more about volumes at:
https://brainly.com/question/13529955
