Question

175. Ifx+y+z=6, (x, y, z>0), then the maximum value of x2yz is (A) 81/4 (B) 27 / 4 (C) 115/4 (D) 91/4

JEE/Engineering Exams

Maths

Solution

136

4.0 (1 ratings)

( x+y+z=6 ) ( frac{x}{2}+frac{x}{2}+y+3=6 ) ( frac{A M geqslant G M}{frac{x}{2}+frac{x}{2}+y+3}>,left(frac{x}{2} cdot frac{x}{2} cdot y cdot 3right)^{1 / 4} ) ( Rightarrow frac{x+y+3}{4} geqleft(frac{x^{2} y z}{2^{2}}right)^{1 / 4} ) ( Rightarrow quad frac{3}{2} quad geqslantleft(frac{x^{2} y partial}{4}right)^{1 / 4} ) ( Rightarrow quad frac{81}{16} geqslant, frac{x^{2} y 3}{4} ) ( Rightarrow quad x^{2} y z leqslant frac{81}{10 x 4} ) option ( A ) Maximum value of ( x^{2} y z=frac{81}{4} )