Relative Minimums and Maximums — Question 9

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Question 9

Consider f(x,y)=x2+y2.f(x,y)=\sqrt{x^2+y^2}.

Tasks

  1. Prove directly that the origin is a strict relative minimum.

  2. Determine whether fx(0,0)f_x(0,0) and fy(0,0)f_y(0,0) exist.

  3. Explain how this example refines the usual critical-point search procedure.

Original worksheet page 1: question and worked solution for 3-3-009
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Question 9 – Solution

Strategy. Use the definition of a relative minimum first, then test each partial derivative with a signed two-sided limit.

Step 1: Direct extremum test At the origin, f(0,0)=0f(0,0)=0. For every (x,y)≠(0,0)(x,y)\ne(0,0), f(x,y)=x2+y2>0=f(0,0).f(x,y)=\sqrt{x^2+y^2}>0=f(0,0). Therefore (0,0) is a strict relative minimum.\boxed{(0,0)\text{ is a strict relative minimum}.}

See the diagram in the original worksheet below.

Step 2: Partial derivatives Along the xx-axis, f(h,0)−f(0,0)h=|h|h,\frac{f(h,0)-f(0,0)}h=\frac{|h|}{h}, whose right-hand limit is 11 and left-hand limit is −1-1. Thus fx(0,0)f_x(0,0) does not exist. The identical calculation on the yy-axis shows that fy(0,0)f_y(0,0) does not exist.

Step 3: Methodological conclusion Fermat’s theorem says that a differentiable interior extremum has zero gradient. It does not say every extremum is differentiable. Consequently, a complete candidate search must include interior points where first partial derivatives vanish and points where they fail to exist. This example supplies the latter case.

Original worksheet page 2: question and worked solution for 3-3-009

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