A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
The side of the square to be cut off is 5 cm.
Explanation
This is an optimization problem from Application of Derivatives. We need to express volume as a function of x (side of square cut), find critical points by setting derivative to zero, and verify maxima using second derivative test. The constraint is that x must be positive and less than half the smaller dimension (24/2 = 12 cm) for the box to exist.
Solution Steps
Step 1: Let x cm be the side of the square cut from each corner. After cutting and folding, the dimensions of the box are: Height = x cm, Length = (45 - 2x) cm, Breadth = (24 - 2x) cm.
Step 2: Volume of the box V(x) = x(45 - 2x)(24 - 2x) = x(1080 - 138x + 4x²) = 4x³ - 138x² + 1080x
Step 3: Differentiating: V'(x) = 12x² - 276x + 1080. Setting V'(x) = 0: 12x² - 276x + 1080 = 0, which gives x² - 23x + 90 = 0. Solving: (x - 5)(x - 18) = 0, so x = 5 or x = 18.
Step 4: Since x must satisfy 0 < x < 12 (from 24 - 2x > 0), we reject x = 18. Thus x = 5 cm.
Step 5: Second derivative test: V''(x) = 24x - 276. At x = 5: V''(5) = 120 - 276 = -156 < 0. Since V''(5) < 0, x = 5 is the point of maxima. Therefore, the side of the square to be cut off should be 5 cm for maximum volume.