Answer:
Step-by-step explanation:
Given
square bottom area is [tex]24\times 24\ in.^2[/tex]
Suppose x in. is cut from each corner to make a open box with maximum volume
New base area is [tex](24-x)\times (24-x) in.^2[/tex]
Volume of box
[tex]V=(24-x)^2\times x[/tex]
Differentiate V w.r.t x to get maximum volume
[tex]\frac{\mathrm{d} V}{\mathrm{d} x}=2\cdot (24-x)(-1)+1\cdot (24-x)^2[/tex]
Put [tex]\frac{\mathrm{d} V}{\mathrm{d} x}=0[/tex]
[tex]\left ( 24-x\right )\left [ -2x+24-x\right ]=0[/tex]
[tex]\left ( 24-x\right )\left [ 24-3x\right ]=0[/tex]
[tex]x=24,8[/tex]
but x=24 is not possible therefore x=8 will yield maximum volume
[tex]V=(24-8)^2\cdot 8=2048\ in.^2[/tex]