1.

If `ab=2a+3b, agt0, b gt0`, then the minimum value of ab isA. 12B. 24C. `(1)/(4)`D. none of these

Answer» Correct Answer - B
`ab=2a+3b rArr b=(2a)/(a-3)`
Now `z=ab=(2a^(2))/(a-3)`
`rArr" "(dz)/(da)=(2[(a-3)2a-a^(2)])/((a-3)^(2))=(2[a^(2)-6a])/((a-3)^(2))`
Put `(dz)/(da)=0, therefore a^(2)-6a=0, a=0,6`
Clearly a = 6 is point of minima
when `a=6, b=4 rArr (ab)_("min")=6xx4=24`


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