## 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.$

**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 $$ D=\{(x,y) \mid 0\leq x\leq 3,0\leq y\leq 2\}. $$

**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.

**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 $$ D=\{(x,y) \mid 0\leq x\leq 3,-3\leq y\leq 0\}. $$

**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 $f(x,y)=\sin x+\cos 2y.$

**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.