Relative Minimums and Maximums — Question 3

Section 16-3 · Calculus pages 3085–3086

Show solutionHide solution

Question 3

Analyze the origin for f(x,y)=x2+4xy+y2.f(x,y)=x^2+4xy+y^2.

Tasks

  1. Verify that the origin is the only critical point.

  2. Classify it using the second derivative test.

  3. Give two explicit paths that make the saddle geometry visible.

Question 3 – Solution

Strategy. The quadratic form itself records the local geometry; the Hessian test and carefully chosen diagonal paths should agree.

Step 1: Critical point fx=2x+4y,fy=4x+2y.f_x=2x+4y,\qquad f_y=4x+2y. Solving x+2y=0x+2y=0 and 2x+y=02x+y=0 gives x=y=0x=y=0. Therefore the origin is the unique critical point.

Step 2: Hessian test fxx=2,fyy=2,fxy=4.f_{xx}=2,\qquad f_{yy}=2,\qquad f_{xy}=4. Hence D=(2)(2)−42=−12<0.D=(2)(2)-4^2=-12<0. Therefore (0,0) is a saddle point.\boxed{(0,0)\text{ is a saddle point}.}

Step 3: Path verification Along y=xy=x, f(x,x)=x2+4x2+x2=6x2>0f(x,x)=x^2+4x^2+x^2=6x^2>0 for x≠0x\ne 0. Along y=−xy=-x, f(x,−x)=x2−4x2+x2=−2x2<0.f(x,-x)=x^2-4x^2+x^2=-2x^2<0.

Diagram shown in the original SVG worksheet.

Both paths approach the origin, whose value is zero, so arbitrarily nearby values occur on both sides of the critical value.