Find two numbers whose sum is 24 and whose product is as large as possible.

Q: Find two numbers whose sum is 24 and whose product is as large as possible.

Explanation: Let one of the numbers be $ x $. Then, the other number is $ 24 - x $. Let the product of the two numbers be: \[ P(x) = x(24 - x) = 24x - x^2 \] Differentiate $ P(x) $: \[ P'(x) = 24 - 2x, \quad P''(x) = -2 \] Set the first derivative to zero for critical points: \[ P'(x) = 0 \Rightarrow 24 - 2x = 0 \Rightarrow x = 12 \] Check the second derivative: \[ P''(12) = -2 < 0 \] Since $ P''(x) < 0 $, the function attains a **maximum** at $ x = 12 $. Therefore, the two numbers are: \[ x = 12, \quad 24 - x = 12 \] The two numbers are 12 and 12. The product is maximum when both are equal.

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Accepted Answer

Explanation: Let one of the numbers be $ x $.  
Then, the other number is $ 24 - x $.

Let the product of the two numbers be:
$$
P(x) = x(24 - x) = 24x - x^2
$$

Differentiate $ P(x) $:
$$
P'(x) = 24 - 2x, \quad P''(x) = -2
$$

Set the first derivative to zero for critical points:
$$
P'(x) = 0 \Rightarrow 24 - 2x = 0 \Rightarrow x = 12
$$

Check the second derivative:
$$
P''(12) = -2 < 0
$$

Since $ P''(x) < 0 $, the function attains a **maximum** at $ x = 12 $.

Therefore, the two numbers are:
$$
x = 12, \quad 24 - x = 12
$$

The two numbers are  12  and  12.  The product is maximum when both are equal.

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