well c it lik this...wrt x y is const.. so lets call it f(x) where coeff include y... now find f'(x)=0 u get rel x=3-y chk for f"(x) also now v get tht @ x=3-y f(x) has min value= 2y^2+4y-9 (which u get by subst x in 1st eqn i.e f(x)) now v have to find min val of this... so diff 2y^2 +4y-9=0 u get min value @ y=-1 so min val is -11 now if u r askin y not i initiall do as f(y) ok do the same way.. u get min value @ 3y=1-x in tht case and if u follow same procedure u again get -11
Moderation Team
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