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Question:

A square tin sheet of side 12 inches is converted into a box with open top in the following steps – the sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box?

Explanation:

When the tin sheet is cut across its corners as shown in the figure, the box formed will have a height of x inches and its base will be a square of side (12 – 2x) inches.

Let the volume of the box, V = (12 – 2x)² × x = 4x³ − 48x² + 144x

For V to be maximum, $\frac{dV}{dx}$ shoule be 0.

i.e. 12x² − 96x + 144 = 0

∴ 12(x − 6)(x − 2) = 0

∴ x = 2 or x = 6

However, x cannot be 6 as the length of the side is (12 – 2x).

∴ x = 2

Hence, option (d).

Alternatively,

Since V = (12 – 2x)² × x = [2(6 – x)]² × x = 4x(6 – x)²,

Substituting values of x from 1 to 5, we get V maximum when x = 2 (i.e. V = 128)