Part 7 - Absolute max and min values
Introduction
In this part we’ll cover how we find absolute maximum and minimum values for a function.
Absolute minimum/maximum values
Let’s start with an example to understand why it’s important to find these values:
A rectangular box is to be made from $12m^2$ of cardboard. Find the maximum volume of such a box.
We can describe the volume of the rectangular box as $V = xyz$.
We also assume we don’t have a top side, therefore the total surface area will be: $$ 2zx + 2zy + yx = 12 $$
Solving for $z$: $$ z = \dfrac{12 - yx}{2(x + y)} $$
We can now write our volume function, as a function of two variables: $$ V(x, y) = \dfrac{xy (12 - yx)}{2(x + y)} = \dfrac{12xy - x^2y^2}{2x + 2y} \newline $$
As we covered last time, we need to solve $\nabla f = 0$, let’s first find the partial derivates: $$ \begin{align*} V_x(x, y) & = \dfrac{f’\cdot g - f \cdot g’}{g^2} \newline & = \dfrac{(12y - 2xy^2)(2x + 2y) - (12xy - x^2y^2)(2)}{(2x + 2y)^2} \newline & = \dfrac{(24xy + 24y^2 - 4x^2y^2 - 4xy^3) - (24xy - 2x^2y^2)}{(2x + 2y)^2} \newline & = \dfrac{24xy + 24y^2 - 4x^2y^2 - 4xy^3 - 24xy + 2x^2y^2}{(2x + 2y)^2} \newline & = \dfrac{24y^2 - 2x^2y^2 - 4xy^3}{(2x + 2y)^2} \newline & = \dfrac{12y^2 - x^2y^2 - 2xy^3}{2(x + y)^2} \newline & = \dfrac{y^2(12 - 2xy - x^2)}{2(x + y)^2} \newline \end{align*} $$
We can see that this equation is symmetrical for $x$ and $y$, therefore: $$ V_y = \dfrac{x^2(12 - 2xy - y^2)}{2(x + y)^2} \newline $$
$$ \begin{cases} 12 - 2xy - y^2 = 0 \newline 12 - 2xy - x^2 = 0 \end{cases} $$
From this we find: $$ y^2 - x^2 = 0 \newline y^2 = x^2 \newline y = x \ | \ \text{since they are only positive} $$
Which leads us to: $$ 12 - 2x^2 - x^2 = 0 \newline 12 - 3x^2 = 0 \newline 3x^2 = 12 \newline x^2 = 4 \newline x = 2 \newline y = 2 \newline $$
Since we cannot get an arbitrarily big volume out of $12m^2$ of cardboard. Since there exists a critical point, this must be the absolute maximum. $$ V_max = V(2, 2) = \ldots = 4 $$
Does absolute minimum/maximum always exist?
Let’s recall for one variable, the so-called extreme variable theorem.
A continuous function on the interval $[a, b]$, always has absolute minimum/maximum.
This is quite intuitive, so let’s not prove it.
Similarly, for functions of two variables, this holds, but we need to define some terminology first.
Bounded set
A bounded set in $\mathbb{R}^2$, is a set that is contained in some disc.
Boundary points
A boundary point at a set $D \in \mathbb{R}^2$, is a point such that, any disc centered at this point contains both some points from $D$ and some points outside $D$.
Closed set
A closed set in $\mathbb{R}^2$, is a set that is contains all of its boundary points
Examples
$$ \{(x, y) \in \mathbb{R}^2 | x^2 + y^2 \leq 1 \} \ | \ \text{contains boundary points, closed set} $$
$$ \{(x, y) \in \mathbb{R}^2 | x^2 + y^2 \le 1 \} \ | \ \text{does not contain boundary points, not closed set} $$
$$ \{(x, y) \in \mathbb{R}^2 | x^2 + y^2 \leq 1, (x, y) \neq (0, 1) \} \ | \ \text{does not contain boundary point, not closed set} $$
$$ \{(x, y) \in \mathbb{R}^2 | y \geq 0 \} \ | \ \text{contains boundary points, closed set} $$
Now we can define the extreme value theorem.
Extreme value theorem
If, $f$, is a continuous function, on a bounded closed region. Then $f$ has absolute values.
How to find absolute values
Recall, for functions of one variable, absolute values are either critical points inside the interval, or at the boundary.
So we need to find the critical points inside $D$ and compute the values. Along with the boundary points, from these computations we can decipher which points are minimums and maximums.
Example
Find absolute minimum and maximum for $f(x, y) = x^2 - 2xy + 2y$, on the rectangle: $$ R = \{(x, y) | \begin{align*} 0 \leq x \leq 3, 0 \leq y \leq 2 \end{align*} \} $$
Since $f$ is a continuous function, and $R$ is bounded closed set. By the extreme value theorem there exists absolute values.
Let’s first investigate critical points inside $R$.
$$ \nabla f = 0 \newline \langle 2x - 2y, -2x + 2 \rangle = 0 $$
$$ \begin{cases} 2x - 2y = 0 \newline -2x + 2 = 0 \end{cases} $$
$$ x = 1 \newline y = 1 $$
Therefore, $(1, 1)$ is a critical point: $$ f(1, 1) = \ldots = \boxed{1} $$
Now we investigate the boundary, points, since this is a rectangle, we can investigate each side. We’ll call these $L_1, L_2, L_3, L_4$: $$ L_1 = \{(x, y) | 0 \leq x \leq 3, y = 0 \} \newline L_2 = \{(x, y) | x = 3, 0 \leq y \leq 2 \} \newline L_3 = \{(x, y) | 0 \leq x \leq 3, y = 2 \} \newline L_4 = \{(x, y) | x = 0, 0 \leq y \leq 2 \} \newline $$
Let’s find the absolute minimum/maximum values on each of these.
On $L_1$: $$ g(x) = f(x, 0) = x^2 \newline g(0) = 0 \ | \ \text{abs. min.} \newline g(3) = 9 \ | \ \text{abs. max.} $$
On $L_2$: $$ g(y) = f(3, y) = 9 - 4y \newline g(2) = 1 \ | \ \text{abs. min.} \newline g(0) = 9 \ | \ \text{abs. max.} $$
On $L_3$: $$ g(x) = f(x, 2) = x^2 - 4x + 4 \newline g(2) = 0 \ | \ \text{abs. min.} \newline g(0) = 4 \ | \ \text{abs. max.} $$
On $L_4$: $$ g(y) = f(0, y) = 2y \newline g(0) = 0 \ | \ \text{abs. min.} \newline g(2) = 4 \ | \ \text{abs. max.} $$
From all these computations we find that: $$ f_{min} = 0 \newline f_{max} = 9 $$