Key Concepts and Formulas

• AM-GM Inequality: For any set of $n$ non-negative real numbers $a_1, a_2, \dots, a_n$, the arithmetic mean is greater than or equal to the geometric mean: $\frac{a_1 + a_2 + \dots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \dots a_n}$ Equality holds if and only if $a_1 = a_2 = \dots = a_n$. This inequality is instrumental in finding the minimum value of a sum when the product of variables is fixed, or vice versa.

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Step-by-Step Solution

Step 1: Analyze the Problem and Identify the Tool We are given that $x, y > 0$ and the constraint $x^3 y^2 = 2^{15}$. We need to find the least value of the expression $3x + 2y$. Since we are looking for the minimum of a sum ($3x+2y$) given a product ($x^3y^2$) and the variables are positive, the AM-GM inequality is the most suitable tool.

Step 2: Strategize the Application of AM-GM To effectively use AM-GM, we need to select terms whose sum is related to $3x+2y$ and whose product is related to $x^3y^2$. The powers in the product $x^3y^2$ are 3 for $x$ and 2 for $y$. This suggests we should consider 3 terms involving $x$ and 2 terms involving $y$. Let's choose the terms as $x, x, x, y, y$. These 5 terms will have a sum of $x+x+x+y+y = 3x+2y$ and a product of $x \cdot x \cdot x \cdot y \cdot y = x^3y^2$.

Step 3: Apply the AM-GM Inequality We apply the AM-GM inequality to the 5 positive terms: $x, x, x, y, y$. The arithmetic mean is $\frac{x+x+x+y+y}{5} = \frac{3x+2y}{5}$. The geometric mean is $\sqrt[5]{x \cdot x \cdot x \cdot y \cdot y} = \sqrt[5]{x^3y^2}$. According to the AM-GM inequality: $\frac{3x+2y}{5} \ge \sqrt[5]{x^3y^2}$

Step 4: Substitute the Given Constraint We are given that $x^3y^2 = 2^{15}$. Substituting this value into the inequality from Step 3: $\frac{3x+2y}{5} \ge \sqrt[5]{2^{15}}$

Step 5: Calculate the Geometric Mean We simplify the term on the right-hand side: $\sqrt[5]{2^{15}} = (2^{15})^{1/5} = 2^{15/5} = 2^3 = 8$

Step 6: Update and Solve the Inequality Substituting the calculated value of the geometric mean back into the inequality: $\frac{3x+2y}{5} \ge 8$ To find the least value of $3x+2y$, we multiply both sides by 5: $3x+2y \ge 5 \times 8$ $3x+2y \ge 40$ This inequality shows that the expression $3x+2y$ must be greater than or equal to 40. Therefore, the least possible value is 40.

Step 7: Check the Condition for Equality The equality in the AM-GM inequality holds if and only if all the terms are equal. In our case, this means $x = y$. We need to check if this condition is consistent with the given constraint $x^3y^2 = 2^{15}$. If $x=y$, then $x^3(x)^2 = 2^{15}$, which simplifies to $x^5 = 2^{15}$. Taking the fifth root of both sides, we get $x = (2^{15})^{1/5} = 2^3 = 8$. So, when $x=8$ and $y=8$, the equality condition is met. Let's verify the value of $3x+2y$ for these values: $3(8) + 2(8) = 24 + 16 = 40$. This confirms that the minimum value of 40 is achievable.

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Common Mistakes & Tips

• Incorrect Term Selection: The most common mistake is not selecting the terms for AM-GM in a way that their product directly relates to the given constraint and their sum relates to the expression to be minimized. For instance, using terms like $3x$ and $2y$ directly in AM-GM would not work due to the powers in the product.
• Ignoring Equality Condition: It is crucial to check if the equality condition ($a_1 = a_2 = \dots = a_n$) can be satisfied under the given constraints. If it cannot, the derived lower bound may not be the actual minimum value.
• Algebraic Errors: Simple calculation mistakes in simplifying roots or solving inequalities can lead to the wrong answer. Double-check all arithmetic operations.

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Summary

This problem is a direct application of the AM-GM inequality. By carefully choosing 5 terms ($x, x, x, y, y$) for the inequality, we were able to relate the sum $3x+2y$ to the given product $x^3y^2$. Applying the AM-GM inequality, substituting the given product, and simplifying led to the inequality $3x+2y \ge 40$. The condition for equality ($x=y=8$) confirms that this lower bound is indeed the least value.

The final answer is $\boxed{40}$ which corresponds to option (D).