Find and classify the critical points of each function I ju

Solution

(i) - (ii) gives 3y - 3x - y² + x² = 0 ===> 3(y-x) - (y-x)(y+x) = 0 ===> (3-y-x)(y-x) = 0 ===> either y+x = 3 or y=x If y=3-x then (ii) becomes 3x-x²-2x(3-x) = 0 ===> x(x-3) = 0 ===> x=0, 3 So two points are ( 0, 3 ) and ( 3, 0 ) If y=x then (ii) becomes 3x-x²-2x² = 0 ===> 3x(1-x) = 0 ===> x=0, 1 So two points are ( 0, 0 ) and ( 1, 1 ) To find the type of stationary point first we find the partial second derivatives F?? = -2y, F?? = -2x, F?? = 3-2x and compute det(H) = F??F??-F??² For each stationary point, we examine these values ( 0, 3 ) : F?? = -6, F?? = 0, F?? = 3, det(H) = -9 Since det(H)<0 this is a saddle point. ( 3, 0 ) : F?? = 0, F?? = -6 , F?? = -3, det(H) = -9 Since det(H)<0 this is a saddle point. ( 0, 0 ) : F?? = 0, F?? = 0 , F?? = 3, det(H) = -9 Since det(H)<0 this is a saddle point. ( 1, 1 ) : F?? = -2, F?? = -2 , F?? = 1, det(H) = 3 Since det(H)>0 and F??<0 this is a maximum with F(1,1) = 1