Local extrema, existence, and constrained optimisation

(a) Critical. $f'_{1}=2x+4y$ and $f'_{2}=4x+2y,$ both $0$ at $(0,0).$

(b) The axes. $f(x,0)=x^{2}>0$ for $x\ne0$ and $f(0,y)=y^{2}>0$ for $y\ne0.$ Along both axes the origin looks like a minimum.

(c) The diagonals. On $y=x{:}$ $f(x,x)=x^{2}+4x^{2}+x^{2}=6x^{2}>0$ for $x\ne0.$ On $y=-x{:}$ $$ \begin{aligned} f(x,-x)&=x^{2}-4x^{2}+x^{2}\\ &=-2x^{2}<0 \end{aligned} $$ for $x\ne0.$ Both lines pass through every disc around the origin, so every such disc contains points where $f$ is larger than $f(0,0)=0$ and points where it is smaller: a saddle point.

What it shows. Two directions (the axes) agreed on "minimum", and a third direction disagreed. That is why the test of the next section needs $B=f''_{12}$ as well as $A$ and $C.$

(a) The critical points. $f'_{1}=3x^{2}-6y=0$ gives $y=\tfrac12x^{2}.$ Then $f'_{2}=3y^{2}-6x=0$ becomes $$ \begin{aligned} \tfrac34x^{4}-6x&=0\,,\\ \tfrac34x\,(x^{3}-8)&=0\,, \end{aligned} $$ so $x=0$ or $x=2,$ with $y=0$ and $y=2.$ The critical points are $(0,0)$ and $(2,2).$

(b) Second derivatives. $f''_{11}=6x,$ $f''_{12}=-6,$ $f''_{22}=6y.$

At $(0,0){:}$ $A=0,$ $B=-6,$ $C=0,$ so $AC-B^{2}=-36<0{:}$ a saddle point, with $f(0,0)=0.$

At $(2,2){:}$ $A=12,$ $B=-6,$ $C=12,$ so $AC-B^{2}=144-36=108>0$ and $A>0{:}$ a strict local minimum point, with $f(2,2)=8+8-24=-8.$

(c) Not global. $f(x,0)=x^{3},$ so for example $f(-3,0)=-27<-8.$ The values of $f$ go below every bound, so $f$ has no global minimum at all, and the local minimum is only local (Watch out 11.7).

(a) $f'_{1}=4x^{3},$ $f'_{2}=2y;$ $g'_{1}=3x^{2},$ $g'_{2}=2y.$ All vanish at $(0,0).$ For $f{:}$ $f''_{11}=12x^{2},$ $f''_{12}=0,$ $f''_{22}=2,$ so at the origin $A=0,$ $B=0,$ $C=2$ and $AC-B^{2}=0.$ For $g{:}$ $g''_{11}=6x,$ $g''_{12}=0,$ $g''_{22}=2,$ the same numbers $0,$ $0,$ $2.$ The test says nothing for either.

(b) For $f.$ $x^{4}\ge0$ and $y^{2}\ge0,$ so $f(x,y)\ge0=f(0,0)$ everywhere, with equality only at the origin: a strict minimum point (global, so certainly local).

(b) For $g.$ On the $x$-axis $g(x,0)=x^{3},$ which is positive for $x>0$ and negative for $x<0.$ Every disc around the origin contains points of both kinds, so the origin is a saddle point (Definition 11.2), even though $g$ curves upwards in the $y$-direction.

The moral. Same $A,$ $B,$ $C,$ different answers: the numbers cannot decide, and you must look at the function.

They exist. $D$ is closed and bounded and $f$ is continuous, so by Fact 11.9 both extreme values are attained.

Interior. $f'_{1}=2x-4=0$ and $f'_{2}=2y=0$ give $(2,0),$ inside $D$ since $4<9.$ There $f(2,0)=4-8=-4.$

Boundary. On $x^{2}+y^{2}=9$ we have $f=9-4x$ with $x\in[-3,3].$ This is decreasing in $x{:}$ largest at $x=-3,$ the point $(-3,0),$ where $f=21;$ smallest at $x=3,$ the point $(3,0),$ where $f=-3.$

Compare. $$ \begin{aligned} f(2,0)&=-4\,,\\ f(-3,0)&=21\,,\\ f(3,0)&=-3\,. \end{aligned} $$ The maximum is $21$ at $(-3,0),$ on the boundary; the minimum is $-4$ at $(2,0),$ inside.

(a) $S_{1}.$ Two weak inequalities between continuous functions, so closed; it lies in the disc of radius $2,$ so bounded. Compact.

(a) $S_{2}.$ Bounded (it lies in the disc of radius $2$), but not closed: the points $(0,y)$ with $0\le y\le1$ are boundary points that the strict inequality $0<x$ leaves out. Not compact.

(a) $S_{3}.$ Closed (three weak inequalities), but not bounded: it contains the whole positive $x$-axis, $(t,0)$ for every $t\ge0.$ Not compact.

(b) $S_{1}.$ Compact, so by Fact 11.9 a maximum exists. Inside, $f'_{1}=f'_{2}=1$ is never $0,$ so the maximum is on the boundary. On the flat edge $y=0,$ $-2\le x\le2,$ $f=x\le2.$ On the arc $x^{2}+y^{2}=4$ the Lagrange conditions $1=2\lambda x,$ $1=2\lambda y$ give $x=y,$ so $x=y=\sqrt2,$ with $f=2\sqrt2\approx2.83.$ The largest value is $2\sqrt2$ at $(\sqrt2,\sqrt2).$

(b) $S_{2}.$ $x\le1$ and $y\le1,$ so $f\le2,$ and $f(1,1)=2$ with $(1,1)\in S_{2}{:}$ the largest value $2$ is attained, although $S_{2}$ is not compact.

(b) $S_{3}.$ $f(t,0)=t$ for every $t\ge0,$ so $f$ takes values larger than any bound: no largest value.

(c) On $S_{2},$ $f>0$ everywhere (since $x>0$), and $f(t,0)=t$ comes as close to $0$ as we like as $t$ shrinks towards $0.$ So $0$ is approached but never attained: no smallest value. Fact 11.9 does not apply on $S_{2},$ and here one extreme exists and the other does not. Its hypotheses are sufficient, not necessary, exactly as Watch out 11.10 says.

The conditions. With $\mathcal{L}=3x+4y-\lambda(x^{2}+y^{2}-25),$ $$ \begin{aligned} 3&=2\lambda x\,,\\ 4&=2\lambda y\,,\\ x^{2}+y^{2}&=25\,. \end{aligned} $$ The first equation rules out $\lambda=0$ and $x=0.$

Eliminate $\lambda.$ Dividing the second equation by the first gives $y/x=4/3,$ so $y=\tfrac43x.$ Then $$ \begin{aligned} x^{2}+\tfrac{16}{9}x^{2}&=25\,,\\ x^{2}&=9\,, \end{aligned} $$ so $x=3,$ $y=4$ or $x=-3,$ $y=-4.$

The candidates. At $(3,4){:}$ $\lambda=3/(2\cdot3)=\tfrac12$ and $f=9+16=25.$ At $(-3,-4){:}$ $\lambda=-\tfrac12$ and $f=-25.$

Why they are the answers. The circle is closed (an equality) and bounded, so compact, and $f$ is continuous: by Fact 11.9 both extremes exist. Here $g'_{1}=2x$ and $g'_{2}=2y$ are never both $0$ on the circle, so by Method 11.14 the extremes are among the candidates. The larger value is the maximum, the smaller the minimum.

(a) The conditions. With $\mathcal{L}=xy^{2}-\lambda(x+2y-m),$ $$ \begin{aligned} y^{2}&=\lambda\,,\\ 2xy&=2\lambda\,,\\ x+2y&=m\,. \end{aligned} $$ The first two give $xy=y^{2}.$ If $y=0$ then $U=0,$ which is not the maximum (the point $x=y=m/3$ gives $U>0$). So $y>0$ and $x=y;$ the constraint gives $3x=m.$ Hence $$ \begin{aligned} x^{*}(m)&=y^{*}(m)=\tfrac{m}{3}\,,\\ \lambda(m)&=y^{2}=\tfrac{m^{2}}{9}\,. \end{aligned} $$

(a) Why the maximum. The budget set with $x,y\ge0$ is a segment from $(m,0)$ to $(0,\tfrac m2){:}$ closed and bounded, and $U$ is continuous, so a maximum exists (Fact 11.9). It is at an endpoint or at a candidate. At both endpoints $U=0,$ while $U(\tfrac m3,\tfrac m3)=\tfrac{m^{3}}{27}>0.$

(b) The value function. $$ \begin{aligned} U^{*}(m)&=\tfrac m3\cdot\bigl(\tfrac m3\bigr)^{2}=\tfrac{m^{3}}{27}\,,\\ \frac{dU^{*}}{dm}&=\tfrac{3m^{2}}{27}=\tfrac{m^{2}}{9}=\lambda(m)\,, \end{aligned} $$ as Fact 11.21 claims.

(c) Estimate against exact. At $m=30{:}$ $U^{*}(30)=1000$ and $\lambda=100.$ So $$ \begin{aligned} U^{*}(31)&\approx1000+100\cdot1\\ &=1100\,,\\ U^{*}(31)&=\tfrac{31^{3}}{27}=\tfrac{29791}{27}\\ &\approx1103.4\,. \end{aligned} $$ The estimate is about $3.4$ too low, about $0.3\%$ of the value: $U^{*}$ is convex in $m,$ so the tangent line lies under it (Watch out 11.23).