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Answer #1Second derivative test for two variables:Suppose that f_(x)(a,b)=0,f_(y)(a,b)=0 and letl=(del^(2))/(delx^(2))f(a,b),m=(del^(2))/(delxdely)f(a,b),n=(del^(2))/(dely^(2))f(a,b)Then (i). f(a,b) is a maximum value if D=ln-m^(2) > 0 and l < 0(ii). f(a,b) is a minimum value if D=ln-m^(2) > 0 and l > 0(iii). f(a,b) is a daddle point. if D=ln-m^(2) < 0(iv). D=ln-m^(2)=0 then f(x,y) fails to have maximum or minimum value and it needs further investigation.Solution:Given function is f(x,y)=7+6x-x^(2)+3y+4y^(2){:[{[f_(x)(x","y)=(del)/(del x)(7+6x-x^(2)+3y+4y^(2))],[f_(y)(x","y)=(del)/(del y)(7+6x-x^(2)+3y+4y^(2))]:}],[=>{[f_(x)(x","y)=6-2x],[f_(y)(x","y)=3+8y]:}],[" and "{[1=f_(xx)(x","y)=-2],[m=f_(xy)(x","y)=0],[n=f_(yy)(x","y)=8]:}]:}Discriminent: D=lm-m^(2)=(-2)(8)-0^(2)=-16Let {[f_(x)(x","y)=6-2x=0],[f_(y)(x","y)=3+8y=0]:} for critical/saddle points.=>{[x=3],[y=-3//8]:}:. critical/saddle points : (x,y)=(3,-(3)/(8))At (x,y)=(3,-(3)/(8))D=1m-m^(2)=(-2)(8)-0^(2)=-16 < 0:.(x,y)=(3,-(3)/(8)) is a saddle point.Seco ... See the full answer