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

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

Explanation: Let one number be $ x $. Then the other number is $ 16 - x $. We want to minimize the sum of their cubes: \[ S(x) = x^3 + (16 - x)^3 \] Differentiate: \[ S'(x) = 3x^2 - 3(16 - x)^2 \] Set $ S'(x) = 0 $ for critical points: \[ 3x^2 = 3(16 - x)^2 \Rightarrow x^2 = (16 - x)^2 \] Expand: \[ x^2 = 256 - 32x + x^2 \Rightarrow -32x + 256 = 0 \Rightarrow x = \frac{256}{32} = 8 \] Now check second derivative: \[ S''(x) = 6x + 6(16 - x) = 6x + 96 - 6x = 96 > 0 \] Since $ S''(8) > 0 $, this confirms a local minimum. \[\] Final Answer: \[ \boxed{\text{The two numbers are } 8 \text{ and } 16-8=8.} \]

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Accepted Answer

Explanation: Let one number be $ x $.  
Then the other number is $ 16 - x $.  
We want to minimize the sum of their cubes:
$$
S(x) = x^3 + (16 - x)^3
$$

Differentiate:
$$
S'(x) = 3x^2 - 3(16 - x)^2
$$

Set $ S'(x) = 0 $ for critical points:
$$
3x^2 = 3(16 - x)^2 \Rightarrow x^2 = (16 - x)^2
$$

Expand:
$$
x^2 = 256 - 32x + x^2 \Rightarrow -32x + 256 = 0
\Rightarrow x = \frac{256}{32} = 8
$$

Now check second derivative:
$$
S''(x) = 6x + 6(16 - x) = 6x + 96 - 6x = 96 > 0
$$

Since $ S''(8) > 0 $, this confirms a local minimum.

Final Answer:
$$
\boxed{\text{The two numbers are } 8 \text{ and } 16-8=8.}
$$

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