Sorry to post the question again. I received the solution from here but there is

Question

Sorry to post the question again. I received the solution from here but there is something wrong with part 3 and 4 of this question. Is (-9,9) the second critical point.....

f(x,y)=5x3+135xy+5y3 fx(x,y)=15x2+135y fy(x,y)=15y2+135x set the first =0, hence: (1/9)x2=y plug into the second to see (15/81)x4+135x=0 x((15/81)x3+135)=0 Hence x=0, -9 When x=0, y=0. When x=-9, y=9 First critical point (0,0) Second (-9,9) Now, consider D(0,0)=fxx(0,0)fyy(0,0)-(fxy(0,0))^2 fxx=30x fyy=30y fxy=135 Then D(0,0)=0-(135)^2<0 Hence the point (0,0) is a saddle point Now, consider D(-9,9)=fxx(-9,9)fyy(-9,9)-(fxy(-9,9))^2 =(-9*30)(9*30)-(135)^2<0 Hence the point (-9,9) is a saddle point

Explanation / Answer

fx(x,y)=15x2+135y therefore y = -(1/9)x^2 not (1/9)x2=y which you have written above therefore the second critical point will be (-9,-9)

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