JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(x, y>0\). If \(x^{3} y^{2}=2^{15}\), then the least value of \(3 x +2 y\) is
- A \(30\)
- B \(32\)
- C \(36\)
- D \(40\)
Answer & Solution
Correct Answer
(D) \(40\)
Step-by-step Solution
Detailed explanation
Using \(A M \geq G M\) \(\frac{x+x+x+y+y}{5} \geq\left(x^{3} \cdot y^{2}\right)^{\frac{1}{5}}\) \(\frac{3 x +2 y }{5} \geq\left(2^{15}\right)^{\frac{1}{5}}\) \((3 x +2 y )_{\min }=40\)
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