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MATH
[1251]

Examples from Global extrema

All examples from Global extrema.

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Extrema on a rectangle

Find the maximum and minimum of

f(x,y)=x2−x2y+3y2f(x,y)=x^2-x^2y+3y^2

on

D={(x,y):−1≤x≤1, −1≤y≤1}.D=\{(x,y):-1\le x\le1,\ -1\le y\le1\}.

For interior candidates,

fx=2x(1−y),fy=−x2+6y.f_x=2x(1-y), \qquad f_y=-x^2+6y.

The equations fx=fy=0f_x=f_y=0 give either x=0x=0, hence y=0y=0, or y=1y=1, hence x2=6x^2=6, which is outside the rectangle. The only interior candidate is (0,0)(0,0), with

f(0,0)=0.f(0,0)=0.

On the vertical edges x=±1x=\pm1,

f(±1,y)=1−y+3y2.f(\pm1,y)=1-y+3y^2.

The derivative is −1+6y-1+6y, so the edge critical point is y=1/6y=1/6, where

f(±1,16)=1−16+3(16)2=1112.f\left(\pm1,\frac16\right) =1-\frac16+3\left(\frac16\right)^2 =\frac{11}{12}.

The edge endpoints give

f(±1,−1)=5,f(±1,1)=3.f(\pm1,-1)=5, \qquad f(\pm1,1)=3.

On the horizontal edges,

f(x,1)=3,f(x,−1)=2x2+3.f(x,1)=3, \qquad f(x,-1)=2x^2+3.

The latter has minimum 33 at x=0x=0 and maximum 55 at x=±1x=\pm1. Comparing all candidates gives

min⁡Df=0 at (0,0){\min_D f=0\text{ at }(0,0)}

and

max⁡Df=5 at (1,−1) and (−1,−1).{\max_D f=5\text{ at }(1,-1)\text{ and }(-1,-1)}.

The minimum is interior, while the maximum occurs at corners.

Extrema on the closed unit disk

Let

f(x,y)=x2+4xy+y2f(x,y)=x^2+4xy+y^2

on

D={(x,y):x2+y2≤1}.D=\{(x,y):x^2+y^2\le1\}.

The interior gradient is

∇f=(2x+4y4x+2y).\nabla f = \begin{pmatrix} 2x+4y\\ 4x+2y \end{pmatrix}.

Solving ∇f=0\nabla f=\mathbf{0} gives (0,0)(0,0), with f(0,0)=0f(0,0)=0.

The boundary is the unit circle. Parameterise it by

x=cos⁡t,y=sin⁡t.x=\cos t, \qquad y=\sin t.

Then

f(cos⁡t,sin⁡t)=cos⁡2t+4cos⁡tsin⁡t+sin⁡2t=1+2sin⁡(2t).\begin{aligned} f(\cos t,\sin t) &=\cos^2t+4\cos t\sin t+\sin^2t\\ &=1+2\sin(2t). \end{aligned}

Since −1≤sin⁡(2t)≤1-1\le\sin(2t)\le1,

−1≤f(cos⁡t,sin⁡t)≤3.-1\le f(\cos t,\sin t)\le3.

The maximum 33 occurs at

(12,12),(−12,−12),\left(\frac1{\sqrt2},\frac1{\sqrt2}\right), \qquad \left(-\frac1{\sqrt2},-\frac1{\sqrt2}\right),

and the minimum −1-1 occurs at

(12,−12),(−12,12).\left(\frac1{\sqrt2},-\frac1{\sqrt2}\right), \qquad \left(-\frac1{\sqrt2},\frac1{\sqrt2}\right).

Comparing with the interior value gives

max⁡Df=3{\max_D f=3}

at the two same-sign boundary points and

min⁡Df=−1{\min_D f=-1}

at the two opposite-sign boundary points.

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