A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
The side of the square to be cut off should be 3 cm for maximum volume.
Explanation
This is an optimization problem using derivatives. We need to express the volume as a function of the side of the cut square, find critical points, and determine the maximum using the second derivative test. The constraint that x cannot exceed half the side length (9 cm) eliminates one critical point.
Solution Steps
Step 1: Let x cm be the side of the square cut from each corner. After cutting, the height of the box = x cm, and the length and breadth = (18 - 2x) cm each.
Step 2: Volume function: V(x) = x(18 - 2x)(18 - 2x) = x(18 - 2x)² = 4x³ - 72x² + 324x
Step 3: First derivative: V'(x) = 12x² - 144x + 324 = 12(x² - 12x + 27) = 12(x - 3)(x - 9)
Step 4: Setting V'(x) = 0 gives x = 3 or x = 9. But x = 9 is not possible as it would make the dimensions zero (18 - = 0). So x = 3 cm.
Step 5: Second derivative test: V''(x) = 24x - 144. At x = 3, V''(3) = 24(3) - 144 = -72 < 0. Since V''(3) < 0, x = 3 is a point of maxima.
Step 6: Therefore, the side of the square to be cut off is 3 cm for maximum volume of the box.