The cut-out from each corner that maximizes the volume is 3 inches.
The dimensions are given as:
[tex]\mathbf{Length = 15}[/tex]
[tex]\mathbf{Width = 24}[/tex]
Assume the cut-out is x.
So, the dimension of the board is:
[tex]\mathbf{Length = 15 - 2x}[/tex]
[tex]\mathbf{Width = 24 - 2x}[/tex]
[tex]\mathbf{Height = x}[/tex]
The volume (V) is calculated as:
[tex]\mathbf{V = Length \times Width \times Height}[/tex]
So, we have:
[tex]\mathbf{V = (15 - 2x) \times (24 - 2x) \times x}[/tex]
Expand
[tex]\mathbf{V = 360x - 78x^2 + 4x^3}[/tex]
Differentiate
[tex]\mathbf{V' = 360 - 156x + 12x^2}[/tex]
Set to 0
[tex]\mathbf{ 360 - 156x + 12x^2 = 0}[/tex]
Divide through by 12
[tex]\mathbf{ 30 - 13x + x^2 = 0}[/tex]
Rewrite as:
[tex]\mathbf{ x^2 - 13x + 30 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = \{3, 10\}}[/tex]
The value of x = 10 is greater than the dimensions of the box.
So, we have:
[tex]\mathbf{x = 3}[/tex]
Hence, the cut-out from each corner that maximizes the volume is 3 inches.
Read more about maximum volumes at:
https://brainly.com/question/7063268
