Let x,y>0. · questionr

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Exams
Joint Entrance Examination
Mathematics
Complex Numbers
Question 159

Question 159 of 455MediumMCQ

Let

x,y>0

. If

x^3y^2=2^{15}

, then the least value of

3x+2y

is

A

30

B

32

C

36

D

40

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Solutions · 1

Official solution

By AM-GM on the five terms

x,x,x,y,y

: \[3x+2y = x+x+x+y+y \ge 5\sqrt[5]{x^3y^2}=5\sqrt[5]{2^{15}}=5\cdot2^3=40,\] with equality when

x=y

(consistent with

x^3y^2=2^{15}

).

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