Answer:
Explanation:
Given a square Piece whose side is 12 inches
Now square pieces are cut from each corner to make it a open box
Suppose x is the length of square piece at each corner
then
base square has a length of [tex]12-2x[/tex]
Dimension of new box is [tex](12-2x)\times (12-2x)\times x[/tex]
Volume [tex]V=(12-2x)\times (12-2x)\times x[/tex]
[tex]V=\left ( 12-2x\right )^2\cdot x[/tex]
For maximum volume differentiate with respect to x we get
[tex]\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0[/tex]
we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid
[tex]V=\left ( 12-2\cdot 4\right )^2\cdot 4[/tex]
[tex]V=4^3[/tex]
[tex]V=64\ in.^3[/tex]
