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Sequences & Series: Minimum Value Positive Real Numbers

JEE Maths question with a full step-by-step solution.

Question
The minimum value of
12x+18y+xy,\frac{12}{x} + \frac{18}{y} + xy ,
 where xx and yy are positive real numbers, is
A12312\sqrt3
B24324\sqrt3
C1818correct
Dinfinity
Solution
Step 1: Notice why AM ≥\ge GM is the right tool - the product of the three terms is a constant:
12x⋅18y⋅xy=12×18=216,\frac{12}{x} \cdot \frac{18}{y} \cdot xy = 12 \times 18 = 216 ,
 with the variables cancelling completely. Step 2: Apply AM ≥\ge GM to the three positive quantities.
12x+18y+xy3≥12x⋅18y⋅xy3=2163=6.\frac{\dfrac{12}{x} + \dfrac{18}{y} + xy}{3} \ge \sqrt[3]{\frac{12}{x}\cdot\frac{18}{y}\cdot xy} = \sqrt[3]{216} = 6 .
 Step 3: Multiply through by 33.
12x+18y+xy≥18.\frac{12}{x} + \frac{18}{y} + xy \ge 18 .
 Step 4: Check that 1818 is actually reached. Equality needs all three terms equal, and since their product is 216216 each must be 2163=6\sqrt[3]{216} = 6:
12x=6  ⟹  x=2,18y=6  ⟹  y=3.\frac{12}{x} = 6 \implies x = 2, \qquad \frac{18}{y} = 6 \implies y = 3 .
 Step 5: Verify the third term agrees: xy=2×3=6xy = 2 \times 3 = 6 . So at (x,y)=(2,3)(x,y) = (2,3) the expression equals 6+6+6=186+6+6 = 18. Answer: (3).
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