Chapter 0 wrap-up

Readiness: what optimization leans on

Closing thought

Every tool in this chapter answers one of three questions about a point. Is it flat? Critical points and the gradient answer that. If it's flat, is it a bottom, a top, or a saddle? The second derivative test answers that, and the eigenvalues of the Hessian carry the same test into any number of variables. If the point has to obey a rule, Lagrange's condition handles an equation, and the corner-point theorem handles straight-line inequalities. The notation, meaning limits, big- $O$ , sums, $x^{(k)}$ and design points, is how the rest of the course will describe a search that improves a guess one step at a time.

You can now find and sort a candidate point by hand. Take $f(x,y)=x^2+2y^2-4x$ . Setting $f_x=2x-4$ and $f_y=4y$ to zero gives $(2,0)$ . There $D=2\cdot4-0=8>0$ and $f_{xx}=2>0$ , so it's a minimum, with $f(2,0)=-4$ . The Hessian's eigenvalues, $2$ and $4$ , are both positive, which says the same thing: a bowl. Next, Chapter 1 gives exact names to what is being chosen, how a choice is judged, and the rules it must obey. Then it sorts problems by whether their pieces are straight or curved, and whether the unknowns must be whole numbers.

Chapter review

One practice set per topic this chapter taught — same fixed, ungraded ladders as the practice panels inside each section.

Practice: Readiness check: critical points, gradients, and Lagrange multipliers (5 problems · doesn’t affect your progress)

For a smooth $f(x,y)$ , what makes an interior point $(a,b)$ a critical point?

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Why

Both first partials have to vanish at once. The value $f(a,b)$ can be anything, and $D$ and $f_{xx}$ belong to the second step, sorting a critical point after it's found.

$\nabla f(a,b)=\mathbf 0$

Find the critical point of $f(x,y)=x^2+xy+y^2-3x$ . What is its $x$ -coordinate?

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Why

$f_y=0$ gives $x=-2y$ . Then $f_x=-4y+y-3=0$ gives $y=-1$ , so $x=2$ . The critical point is $(2,-1)$ .

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Why

Set $f_x=2x+y-3$ and $f_y=x+2y$ both to zero.

For $f(x,y)=x^2+xy+y^2-3x$ at its critical point $(2,-1)$ , what does the second derivative test say?

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Why

$f_{xx}=2$ , $f_{yy}=2$ , $f_{xy}=1$ , so $D=4-1=3>0$ . With $f_{xx}>0$ the surface curves up, giving a minimum. A nonzero $f_{xy}$ only matters through $D$ .

A minimum, since $D=3>0$ and $f_{xx}>0$ .

For $f(x,y,z)=xyz+z^2$ , what is the third component of $\nabla f$ at $(1,2,3)$ ?

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Why

$f_z=xy+2z$ , which is $2+6=8$ at $(1,2,3)$ . The full gradient there is $(6,3,8)$ .

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Maximize $f(x,y)=xy$ on the line $x+y=10$ using $\nabla f=\lambda\nabla g$ with $g(x,y)=x+y$ . What is $\lambda$ ?

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Why

$(y,x)=\lambda(1,1)$ gives $x=y=\lambda$ . Then $x+y=10$ gives $x=y=5$ , so $\lambda=5$ and the maximum is $f=25$ .

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Why

$\nabla f=(y,x)$ and $\nabla g=(1,1)$ .

Practice: Readiness check: eigenvalues, quadratic forms, corner points, and design points (5 problems · doesn’t affect your progress)

The $2\times3$ matrix $A=\begin{pmatrix}1&0&2\\0&1&-1\end{pmatrix}$ acts on the design point $\mathbf x=(3,1,2)$ in $\mathbb R^3$ . What is the first entry of $A\mathbf x$ ?

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Why

The first row times $\mathbf x$ : $1\cdot3+0\cdot1+2\cdot2=7$ . The full product is $(7,-1)$ , a vector in $\mathbb R^2$ : a $2\times3$ matrix turns a 3-component point into a 2-component one.

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Why

What is the smaller eigenvalue of $A=\begin{pmatrix}6&2\\2&3\end{pmatrix}$ ?

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Why

$(6-\lambda)(3-\lambda)-4=\lambda^2-9\lambda+14=(\lambda-7)(\lambda-2)$ . The eigenvalues are $7$ and $2$ .

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Why

The characteristic polynomial is $\lambda^2-9\lambda+14$ .

The symmetric matrix $A=\begin{pmatrix}1&3\\3&1\end{pmatrix}$ has eigenvalues $4$ and $-2$ . What kind of quadratic form is $Q(\mathbf x)=\mathbf x^{\mathsf T}A\mathbf x$ ?

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Why

One positive and one negative eigenvalue means $Q$ rises in one direction and falls in another. Positive entries don't decide it: $Q(1,-1)=1-6+1=-4<0$ .

Indefinite: a saddle, since the eigenvalues have opposite signs.

A shaded region has corners $(0,0)$ , $(5,0)$ , $(3,4)$ , and $(0,6)$ . What is the largest value of $P=5x+4y$ on the region?

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Why

Check each corner: $0$ , $25$ , $15+16=31$ , and $24$ . The corner-point theorem says the best value is at a corner, so the maximum is $31$ , at $(3,4)$ .

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Why

For the same region and $P=5x+4y$ , why can't the best point lie strictly inside the shaded region?

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Why

A small step $t(5,4)$ raises $P$ by $41t$ and, from an inside point, stays in the region. So no inside point can be best. You keep improving until you hit the edge, and then slide along it to a corner.

Ask what happens to $P$ if you nudge an inside point by a small step $t(5,4)$ .

From any inside point, a small step along $(5,4)$ still raises $P$ .

Practice: Readiness check: limits, big-O, sums, and iterate sequences (5 problems · doesn’t affect your progress)

As $k$ grows without bound, what number does the ratio $\frac{5k-1}{2k+3}$ settle toward?

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Why

Divide top and bottom by $k$ : $\frac{5-1/k}{2+3/k}\to\frac52=2.5$ .

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Why

2.5

Is $n^2=O(n)$ ?

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Why

Big- $O$ needs one fixed $c$ that works for every large $n$ . Any $c$ fails once $n>c$ . Agreeing at a single point, or both growing, isn't enough.

No: $n^2/n=n$ is unbounded, so no constant $c$ works.

For $f(n)=4n^2+5n+1$ , what is the smallest whole number $c$ with $f(n)\le c\,n^2$ for every $n\ge5$ ?

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Why

The left side is largest at $n=5$ : $4+1+0.04=5.04$ . So $c=5$ is too small and $c=6$ works. Starting at a later $n$ lets you use a smaller $c$ than you'd need from $n=1$ .

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Divide by $n^2$ : you need $4+\frac5n+\frac1{n^2}\le c$ . Where is the left side largest for $n\ge5$ ?

Compute $\displaystyle\sum_{j=0}^{3}2^j$ .

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Why

$2^0+2^1+2^2+2^3=1+2+4+8=15$ . The index starts at $0$ , so there are four terms, not three.

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Why

The iterates follow $x^{(k+1)}=0.5\,x^{(k)}+3$ with $x^{(0)}=2$ . What value do they approach as $k\to\infty$ ?

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Why

The iterates go $2,4,5,5.5,\ldots$ , and the gap to $6$ halves each step. A limit $L$ must satisfy $L=0.5L+3$ , which gives $L=6$ .

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Why

Write out $x^{(1)}$ , $x^{(2)}$ , $x^{(3)}$ , or ask which value the rule leaves unchanged.

Take the chapter quiz On to Chapter 1 →

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