2.10 Q-3

Question Statement

Find two positive integers whose sum is 12 and the product of one integer with the square of the other is maximized.

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Background and Explanation

This problem requires us to find two positive integers, $x$ and $12 - x$, whose sum is fixed at 12, and we want to maximize the product of one integer with the square of the other. To solve this, we will use calculus to find the critical points of the function that represents the product and determine the maximum by applying the second derivative test.

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Solution

1. Define the variables:

   Let the two integers be $x$ and $12 - x$.

2. Set up the product function:

   We want to maximize the product of $x$ and the square of $12 - x$. The function for the product is:

$$f(x) = x(12 - x)^2$$

3. Differentiate the function:

   Now, we differentiate $f(x)$ with respect to $x$. Using the product rule, we get:

$$f'(x) = (12 - x)^2 + x \cdot 2(12 - x)(-1)$$

Simplifying:

$$f'(x) = (12 - x)^2 - 2x(12 - x)$$

Factor out $(12 - x)$:

$$f'(x) = (12 - x)[(12 - x) - 2x]$$

Simplifying further:

$$f'(x) = (12 - x)(12 - 3x)$$

4. Set the derivative equal to zero:

   To find the critical points, set $f'(x) = 0$:

$$(12 - x)(12 - 3x) = 0$$

5. Solve for $x$:

   Solve the two factors separately:

   • $12 - x = 0 \Rightarrow x = 12$ (but this is not a valid solution since $12 - x$ must be positive)
   • $12 - 3x = 0 \Rightarrow x = 4$

   So, $x = 4$ is the valid solution.

6. Check the second derivative:

   To confirm that this value of $x$ gives a maximum, we compute the second derivative:

$$f''(x) = \frac{d}{dx}[3(12 - x)(4 - x)] = -3(16 - 2x)$$

Simplifying:

$$f''(x) = -6(8 - x)$$

When $x = 4$:

$$f''(4) = -6(8 - 4) = -24$$

Since $f''(x)$ is negative, this confirms that $x = 4$ gives a maximum.

7. Find the two integers:

   The first integer is $x = 4$, and the second integer is $12 - x = 8$.

Thus, the two integers are 4 and 8, and the product of one with the square of the other is maximized.

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Key Formulas or Methods Used

• Product function: $f(x) = x(12 - x)^2$
• First derivative: $f'(x) = (12 - x)(12 - 3x)$
• Second derivative test: $f''(x) = -6(8 - x)$

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Summary of Steps

1. Define the two integers $x$ and $12 - x$.
2. Set up the product function: $f(x) = x(12 - x)^2$.
3. Differentiate the function to get $f'(x) = (12 - x)(12 - 3x)$.
4. Set the derivative equal to zero: $(12 - x)(12 - 3x) = 0$, solving gives $x = 4$.
5. Check the second derivative to confirm a maximum.
6. The two integers are $x = 4$ and $12 - 4 = 8$, which maximize the product.