Question

A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bending upward the projecting portion which remain. Find the maximum volume that can be obtained.

Answer 1

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 $12-2x$

Dimension of new box is $(12-2x)\times (12-2x)\times x$

Volume $V=(12-2x)\times (12-2x)\times x$

$V=\left ( 12-2x\right )^2\cdot x$

For maximum volume differentiate with respect to x we get

$\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0$

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

$V=\left ( 12-2\cdot 4\right )^2\cdot 4$

$V=4^3$

$V=64\ in.^3$