Respuesta :

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]    

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