Question 53 of 82intermediate🔧 ApplyNumerical4 marks

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Correct Answer

The two positive numbers are 8 and 8.

Exercise: EXERCISE 6.3 | Q: 16 | (Page 30)
For More Understanding

Explanation

This question follows the same pattern as Example 22 in the textbook, which finds two numbers whose sum is fixed and whose squares sum is minimum. Here, we adapt the method for cubes instead of squares. The key insight is that when minimizing a symmetric function of two numbers with a fixed sum, the optimal solution often occurs when both numbers are equal.

Solution Steps

  1. Step 1: Let one number be x. Since the sum is 16, the other number is (16 - x).

  2. Step 2: Let S(x) denote the sum of cubes: S(x) = x³ + (16 - x)³

  3. Step 3: Find the first derivative: S'(x) = 3x² - 3(16 - x)²

  4. Step 4: Set S'(x) = 0: 3x² - 3(16 - x)² = 0, which gives x² = (16 - x)². Solving: x = 16 - x, so x = 8.

  5. Step 5: Find the second derivative: S''(x) = 6x + 6(16 - x) = 96

  6. Step 6: Since S''(8) = 96 > 0, x = 8 is a point of local minimum. Hence, the two numbers are 8 and 16 - 8 = 8.

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