Relative Minimums and Maximums — Question 7

Section 16-3 · Calculus pages 3093–3094

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Question 7

A twice differentiable function has the local expansion near a critical point PP: f(P+(h,k))=5+3h2+4hk−k2+o(h2+k2).f(P+(h,k))=5+3h^2+4hk-k^2+o(h^2+k^2).

Tasks

  1. Reconstruct the Hessian matrix at PP.

  2. Apply the second derivative test.

  3. Use the remainder term carefully to prove the classification along two coordinate directions.

Question 7 – Solution

Strategy. Match the quadratic terms with 12(h,k)H(h,k)T\tfrac 12(h,k)H(h,k)^{T}, then show that the leading signs dominate the smaller remainder.

Step 1: Hessian The second-order Taylor term is 12(fxxh2+2fxyhk+fyyk2).\frac 12\left(f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\right). Matching coefficients gives fxx=6,fxy=4,fyy=−2,f_{xx}=6,\qquad f_{xy}=4,\qquad f_{yy}=-2, so H(P)=(644−2).\boxed{H(P)=\begin{pmatrix}6&4\\4&-2\end{pmatrix}}.

Step 2: Second derivative test D=(6)(−2)−42=−28<0.D=(6)(-2)-4^2=-28<0. Therefore PP is a saddle point.

Step 3: Remainder verification Along k=0k=0, f(P+(h,0))−5=3h2+o(h2),f(P+(h,0))-5=3h^2+o(h^2), which is positive for all sufficiently small nonzero hh. Along h=0h=0, f(P+(0,k))−5=−k2+o(k2),f(P+(0,k))-5=-k^2+o(k^2), which is negative for all sufficiently small nonzero kk. The little-oo terms are eventually smaller in magnitude than, for example, half of the displayed quadratic terms. Thus nearby values occur on both sides of 55, rigorously confirming the saddle.