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Week 9 Discussion Notes

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Second Partials Test

If f(x,y)f\p{x, y} is a twice differentiable function, then a point (a,b)\p{a, b} is a critical point if ∇f(a,b)=0⃗\nabla f\p{a, b} = \vec{0}. The discriminant of ff at (a,b)\p{a, b} is

D(a,b)=fxx(a,b)fyy(a,b)−fxy2(a,b).D\p{a, b} = f_{xx}\p{a, b} f_yy\p{a, b} - f_{xy}^2\p{a, b}.

With the definitions out of the way, we have this nice result that helps you classify critical points:

Theorem (second partials test)

Assume that (a,b)\p{a, b} is a critical point of f(x,y)f\p{x, y}. Then we have the following chart:

D(a,b)fxx(a,b)classification>0>0local minimum>0<0local maximum<0saddle point=0inconclusive\begin{array}{ccl} \\[-4ex]\hline\\[-2ex] D\p{a, b} & f_{xx}\p{a, b} & \text{classification} \\[1ex]\hline\\[-1.75ex] > 0 & > 0 & \text{local minimum} \\ > 0 & < 0 & \text{local maximum} \\ < 0 & & \text{saddle point} \\ = 0 & & \text{inconclusive} \\[1ex]\hline \end{array}
Remark.

When D(a,b)>0D\p{a, b} > 0, you can check the sign of either fxx(a,b)f_{xx}\p{a, b} or fyy(a,b)f_{yy}\p{a, b} to figure out if it's a local minimum or local maximum.

Also, if you're curious, the discriminant is

D(a,b)=det⁡(fxx(a,b)fxy(a,b)fyx(a,b)fyy(a,b))D\p{a, b} = \det \begin{pmatrix} f_{xx}\p{a, b} & f_{xy}\p{a, b} \\ f_{yx}\p{a, b} & f_{yy}\p{a, b} \end{pmatrix}

and the matrix above is called the Hessian of ff at (a,b)\p{a, b}.

Optimization

Generally, if you want to optimize a function f(x,y)f\p{x, y} on some domain DD, there are two steps:

  1. Optimize on the boundary of DD.
    • You can do this by parametrizing the boundary with r⃗(t)\vec{r}\p{t}, and then optimizing the 1D function f(r⃗(t))f\p{\vec{r}\p{t}}.
    • Later, you'll learn that you can do this with Lagrange multipliers.
  2. Optimize on the interior of DD.
    • To do this, you'll want to find the critical points of ff that lie in DD.
Example 1.

Optimize f(x,y)=x2+y2+xf\p{x, y} = x^2 + y^2 + x on the domain x2+y2≤1x^2 + y^2 \leq 1, and classify the critical points of ff in this interior of the domain.
Solution.

Let's follow the two steps. The domain here is the disk of radius 11, so its boundary is the unit circle.

Boundary:

To parametrize the unit circle, we can use r⃗(t)=⟨cos⁡t,sin⁡t⟩\vec{r}\p{t} = \ang{\cos{t}, \sin{t}}. Then we need to optimize

f(r⃗(t))=cos⁡2t+sin⁡2t+cos⁡t=1+cos⁡t.f\p{\vec{r}\p{t}} = \cos^2{t} + \sin^2{t} + \cos{t} = 1 + \cos{t}.

cos⁡t\cos{t} is maximized when t=0t = 0 and minimized when t=πt = \pi, so these give the critical points

r⃗(0)=⟨1,0⟩andr⃗(π)=⟨−1,0⟩.\vec{r}\p{0} = \ang{1, 0} \quad\text{and}\quad \vec{r}\p{\pi} = \ang{-1, 0}.

Interior:

For the interior, we'll want to use the second partials test. The gradient of ff is

∇f(x,y)=⟨2x+1,2y⟩=⟨0,0⟩.\nabla f\p{x, y} = \ang{2x + 1, 2y} = \ang{0, 0}.

This gives the critical point (a,b)=(−12,0)\p{a, b} = \p{-\frac{1}{2}, 0}.

To find the global minimum and maximum, we just need to try all the critical points:

f(1,0)=2f(−1,0)=0f(−12,0)=−14,\begin{aligned} f\p{1, 0} &= 2 \\ f\p{-1, 0} &= 0 \\ f\p{-\frac{1}{2}, 0} &= -\frac{1}{4}, \end{aligned}

so the global minimum is −14-\frac{1}{4} at (−12,0)\p{-\frac{1}{2}, 0} and the global maximum is 22 at (1,0)\p{1, 0}.

Lastly, the problem wants us to classify the critical point (−12,0)\p{-\frac{1}{2}, 0}. We just need to calculate the discriminant:

fxx(x,y)=2fyy(x,y)=2fxy(x,y)=0.\begin{aligned} f_{xx}\p{x, y} &= 2 \\ f_{yy}\p{x, y} &= 2 \\ f_{xy}\p{x, y} &= 0. \end{aligned}

so for this problem,

D(−12,0)=4>0andfxx(−12,0)=2>0.D\p{-\frac{1}{2}, 0} = 4 > 0 \quad\text{and}\quad f_{xx}\p{-\frac{1}{2}, 0} = 2 > 0.

Thus, by the second partials test, (−12,0)\p{-\frac{1}{2}, 0} is a local minimum of ff.