Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesMeans / inequalityMedium2 minPYQ_2022

MathematicsMediumstatement

Let $x, y > 0$ . If $x^{3} y^{2} = 2^{15}$ , then the least value of $3 x + 2 y$ is

Answer:

Solution:

Using $A . M \geq G . M$ , we get

$\frac{x + x + x + y + y}{5} \geq \sqrt[5]{x^{3} \cdot {(y)}^{2}}$

$\Rightarrow \frac{3 x + 2 y}{5} \geq \sqrt[5]{x^{3} y^{2}}$

$\Rightarrow \frac{3 x + 2 y}{5} \geq \sqrt[5]{(2^{15})} (∵ x^{3} y^{2} = 2^{15})$

$\Rightarrow 3 x + 2 y \geq 5 \cdot {(2)}^{15 / 5}$

$\Rightarrow 3 x + 2 y ⩾ 5 \times 2^{3}$

$\Rightarrow 3 x + 2 y ⩾ 40$

$∴$ The minimum value of $3 x + 2 y$ is $40$

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Means / inequality

⏱ 2mℹ️ Source: PYQ_2022

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