Absolute Extrema (and the Extreme Value Theorem)

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

Absolute Maximum and Absolute Minimum

Definition. Let $f$ be a function of two variables $x$ and $y.$

(1) The function $f$ has an absolute maximum at $\left(x_0,y_0\right)$ if $f(x,y)\leq f\left(x_0,y_0\right)$ for all $(x,y)$ in the domain $D$ of $f.$

(2) The function $f$ has an absolute minimum at $\left(x_0,y_0\right)$ if $f(x,y)\geq f\left(x_0,y_0\right)$ for all $(x,y)$ in the domain $D$ of $f.$

Collectively, absolute maxima and absolute minima are called absolute extrema. When does a function have any absolute extrema?

Extreme Value Theorem

Theorem. (Extreme Value) A function $f$ attains both an absolute maximum and an absolute minimum on any closed bounded set $S$ where it is continuous.

Example. Find the shortest distance from the point $(1,0,-2)$ to the plane $ x+2y+z=4.$

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The shortest distance from the point $(1,0,-2)$ to the plane $x+2y+z=4$ is $5\sqrt{6}/6.$

Solution. The distance from any point $(x,y,z)$ to the point $(1,0,-2)$ is \begin{equation} d=\sqrt{(x-1)^2+y^2+(z+2)^2}\end{equation} but if $(x,y,z)$ lies on the plane, then $z=4-x-2y$ and so we have \begin{equation}d=\sqrt{(x-1)^2+y^2+(6-x-2y)^2}.\end{equation} We can minimize $d$ by minimizing the simpler expression \begin{equation}d^2=f(x,y)=(x-1)^2+y^2+(6-x-2y)^2.\end{equation} By solving the equations \begin{equation}f_x=2(x-1)^2-2(6-x-2y)=4x+4y-14=0\end{equation} \begin{equation}f_y=2y-4(6-x-2y)=4x+10y-24=0\end{equation} we find that the only critical point is $(11/3,5/3).$ Since $f_{x x}=4,$ $f_{x y}=4,$ and $f_{y y}=10,$ we have \begin{equation}D(x,y)=f_{x x}f_{y y}-\left(f_{x y}\right){}^2=24>0\end{equation} and $f_{x x}>0$ so $f$ has a local minimum at $(11/6,5/3).$ Intuitively, we can see that this local minimum is actually an absolute minimum because there must be a point on the given plane that is closest to $(1,0,-2).$ If $x=11/6$ and $y=5/3,$ then \begin{equation} d =\sqrt{\left(\frac{5}{6}\right)^2 + \left(\frac{5}{3}\right)^2 +\left(\frac{5}{6}\right)^2} =\frac{5\sqrt{6}}{6}.\end{equation} which is the shortest distance.

Example. Find the absolute extrema of the function $ f(x,y)=x^2-2x y+2y $ over the bounded region over the rectangle $$ D=\{(x,y) \mid 0\leq x\leq 3,0\leq y\leq 2\}. $$

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Finding absolute extrema of $f(x,y)=x^2-2x y+2y$ on $D=\{(x,y) \mid 0\leq x\leq 3,0\leq y\leq 2\}.$

Solution. Since $f$ is a polynomial it is continuous on the closed bounded rectangle $D,$ therefore $f$ has both absolute maximum and minimum values. We first find the critical points by solving the system \begin{equation}f_x=2x-2y=0 \qquad \text{and} \qquad f_y=-2x+2=0\end{equation} The only critical point is $(1,1),$ and the value of $f$ there is $f(1,1)=1.$ We look at the values of $f$ on the boundary of $D$, which consists of four line segments $L_1, L_2, L_3$, and $L_4$ as follows.

  • On $L_1$ we have $y=0$ and $f(x,0)=x^2$ for $0\leq x\leq 3.$ This is an increasing function of $x,$ so its minimum value is $f(0,0)=0$ and its maximum value is $f(3,0)=9.$
  • On $L_2$ we have $x=3$ and $f(3,y)=9-4y$ on $0\leq y\leq 2.$ This is a decreasing function of $y,$ so its maximum value is $f(3,0)=9$ and its minimum value is $f(3,2)=1.$
  • On $L_3$ we have $y=2$ and $f(x,2)=x^2-4x+4$ on $0\leq x\leq 3.$ By observing that $f(x,2)=(x-2)^2,$ we see that the minimum value of this function is $f(2,2)=0$ and the maximum value is $f(0,2)=4.$
  • On $L_4$ we have $x=0$ and $f(0,y)=2y$ on $0\leq y\leq 2$ with maximum value $f(0,2)=4$ and minimum value $f(0,0)=0.$

Thus on the boundary, the minimum value of $f$ is 0 and the maximum is 9. We compare these values with the value $f(1,1)=1$ at the critical point and conclude that the absolute maximum value of $f$ on $D$ is $f(3,0)=9$ and the absolute minimum value is $f(0,0)=f(2,2)=0.$

Example. A rectangular box without a lid is to be made from 12 $m^2$ of cardboard. Find the maximum volume of such a box.

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Rectangular box enclosing maximum volume constructed from 12 $\text{m}^2$ of material.

Solution. Let the length, width, and height of the box (in meters) be $x,$ $y,$ and $z.$ Then the volume of the box is $V=x y z.$ We can express $V$ as a function of just two variables by using the fact that the surface area of the sides and the bottom of the box is $2x z+2 y z+x y=12.$ Solving these equation for $z,$ we get \begin{equation} z=\frac{12-x y}{2(x+y)}, \end{equation} so the expression for volume $V$ becomes \begin{equation} V=\frac{12 x y-x^2y^2}{2(x+y)}. \end{equation} We compute the partial derivatives \begin{equation} V_x=\frac{y^2\left(12-2 x y-x^2\right)}{2(x+y)} \qquad \text{and}\qquad V_y=\frac{x^2\left(12-2 x y -y^2\right)}{2(x+y)^2}. \end{equation} If $V$ is a maximum, then $V_x=V_y=0,$ but $x=0$ or $y=0$ gives $V=0,$ so we must solve the equations \begin{equation} 12-2x y-x^2=0\qquad \text{and} \qquad 12-2 x y-y^2=0. \end{equation} These imply that $x^2=y^2$ and so $x=y.$ (Note that $x$ and $y$ must both be positive in this example.) If we put $x=y$ in either equation we get $12-3x^2=0,$ which gives $x=2,$ $y=2,$ and $z=1.$ From the physical nature of this example there must be an absolute maximum volume that has to occur at a critical point of $V,$ so it must be when $x=2,$ $y=2,$ and $z=1.$ Then $V=2\cdot 2\cdot 1=4,$ so the maximum volume of the box is $4m^3.$

Example. Find the absolute extrema of the function $ f(x,y)=x^2+3y^2-4x+2y-3$ over the rectangle $$ D=\{(x,y) \mid 0\leq x\leq 3,-3\leq y\leq 0\}. $$

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Finding absolute extrema of $f(x,y)=x^2+3y^2-4x+2y-3$ on $D=\{(x,y) \mid 0\leq x\leq 3,-3\leq y\leq 0\}.$

Solution. We compute $f_x=2x-4$ and $f_y=6y+2$ and set $f_x=f_y=0$ and find that $(2,-1/3)$ is the only critical point in the interior.

  • On $x=t, y=0$ for $0\leq t\leq 3$, we have $F_1(t)=f(t,0)=t^2-4t-3=0 $ and so $F_1′(t)=2t-4$ yielding the point $(2,0).$
  • On $x=3, y=t$ for $-3\leq t\leq 0$, we have $F_2(t)=f(3,t)=3t^2+2t-6 $ and so $F_2′(t)=6t+2=0$ yielding the point $(3,-1/3).$ On $x=t, y=-3$ for $0\leq t\leq 3$, we have $$ F_3(t)=f(t,-3)=t^2-4t+18 $$ and so $F_3′(t)=2t-4=0$ yielding the point $(2,-3).$
  • On $x=0, y=t$ for $-3\leq t\leq 0$, we have $F_4′(t)=f(0,t)=3t^2+2t-3 $ and so $F_4′(t)=6t+2=0$ yielding the point $(0,-1/3).$

Finally, we have $f(2,-1/3)=-22/3$ (the minimum), $f(3,-1/3)=-19/3,$ $f(2,-3)=14,$
$f(0,-1/3)=-10/3,$ $f(2,0)=-7,$ $f(0,0)=-3,$ $f(3,0)=-6,$ $f(3,-3)=15,$ and $f(0,-3)=18$ (the maximum).

Example. Find the hottest and coldest points on the metal plate given as the region $R=[0,\pi ]\times [0,\pi ],$ whose temperature is given by $f(x,y)=\sin x+\cos 2y.$

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Finding absolute extrema of $f(x,y)=\sin x+\cos 2y$ on $R=[0,\pi ]\times [0,\pi ].$

Solution. Since $f$ is continuous and $R$ is closed and bounded, we know that the absolute maximum and minimum exist. We find that $$ \nabla f=(\cos x) i+(-2\sin 2y) j $$ and so the only critical point in the interior of $R$ is $(\pi /2,\pi /2).$ This is a saddle point because the discriminant of $f$ at $(\pi / 2,\pi /2)$ is negative. The boundary of $R$ consists of four line segments $L_1,$ $L_2,$ $L_3,$ and $L_4$ as follows.

On $L_1$ and $L_3,$ we have $$ f(x,0)=f(x,\pi) =\sin x+1 \quad \text{ for } 0\leq x\leq \pi $$ which achieves a maximum at $\pi /2$ and a minimum at $0$ and $\pi .$ Similarly, on $L_2$ and $L_4$ we have $$ f(0,y)=f(\pi ,y) =\cos 2y \quad \text{ for } 0\leq y\leq \pi $$ which achieves its maximum at $0$ and $\pi $ and its minimum at $\pi /2.$ We see that the hottest points are $(\pi /2,0)$ and $(\pi /2,\pi )$ and the coldest points are $(0,\pi /2)$ and $(\pi ,\pi /2).$

Exercises on Absolute Extrema

Exercise. Find the absolute extrema for the function $f(x,y)=x y^2-2xy+3y$ in the triangular domain with vertices $(0,0),$ $(1,0),$ and $(1,1).$

Exercise. Find the absolute maximum and minimum values for each of the following functions.

$(1) \quad f(x,y)=e^{x^2+2x+y^2}$ on the disk $x^2+2x+y^2\leq 0.$

$(2) \quad f(x,y)=x^2+x y +y^2$ on the disk $x^2+y^2\leq 1.$

$(3) \quad f(x,y)=x y-2x-5y$ on the triangular region $S$ with vertices $(0,0),$ $(7,0),$ and $(7,7).$

$(4) \quad f(x,y)=x^2-4 x y+y^3+4y$ on the square $0\leq x\leq 2,$ $0\leq y\leq 2.$

$(5) \quad f(x,y)=x^2+3y^2-4x+2y-3$ on the square region $S$ with vertices $(0,0),$ $(3,0),$ $(3,-3),$ and $(0,-3).$

Exercise. Find three positive numbers whose sum is 54 and whose product is as large as possible.

Exercise. A wire of length $L$ is cut into three pieces that are bent to form a circle, a square, and a equilateral triangle. How should the cuts be made to minimize the sum of the total area.?

Exercise. A rectangular box with no top is to have a fixed volume. What should its dimensions be if we want to use the least amount of material in its construction?

Exercise. Let $R$ be the triangular region in the $x y$-plane with vertices $(-1,-2),$ $(-1,2),$ and $(3,2).$ A plate in the shape of $R$ is heated so that the temperature at $(x,y)$ is $$ T(x,y)=2x^2-x\text{ }y+y^2-2y+1 $$ (in degrees Celsius). At what point in $R$ or on its boundary is $T$ maximized? What are the temperatures?

Exercise. Find the maximum and minimum values for the given function in the given closed region.

$(1) \quad z=8x^2+4y^2+4y+5$, $x^2+y^2\leq 1$

$(2) \quad z=6x^2+y^3+6y^2$, $x^2+y^2\leq 25$

$(3) \quad z=8x^2-24xy+y^2$, $x^2+y^2\leq 25$

Exercise. Find the volume of the largest box that can be inscribed in the ellipsoid $$
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1. $$

Exercise. Assuming that the function $$ F= 2x^2+6y^2+45z^2-4xy+6yz+12xz-6y+14 $$ has a minimum, find it. Prove that this function has no maximum value.