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15.7 Max/min problems (part 2)

If \( y = f(x) \), \( a \leq x \leq b \) then the absolute max/min of \( f(x) \) can be at the ends of the interval or at critical pts inside the interval.

Figure: 2D plot of f(x) from x=a to x=b with labeled absolute minimum at a valley and absolute maximum at endpoint b.

Procedure:

Find all interior critical pts
Compare \( f(x) \) at the critical pts and at ends to find max/min

\( z = f(x, y) \) is a surface and domain is a region in xy-plane and the boundaries are curves.

The basic idea is the same:

Find all interior critical pts
Then find all possible locations of max/min on the boundaries
Then compare \( f(x, y) \)

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Example Problem

Find max/min of

\[ z = f(x, y) = (x-1)^2 + (y-2)^2 \]

on the region defined by:

\( 0 \leq x \leq 3 \)
\( 0 \leq y \leq 3 \)

Figure: 3D plot of a paraboloid surface opening upwards, centered at (1, 2), restricted to a rectangular domain.

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Contour Map Analysis

Find max/min of

\[ z = f(x, y) = (x-1)^2 + (y-2)^2 \text{ on } 0 \leq x \leq 3, 0 \leq y \leq 3 \]

Figure: Contour map with circular level curves z=2, 6, 8 and a central point z=0, overlaid with a rectangular boundary.

Boundary Point:

\( x=0, y=3 \) (Level 2)

Boundary Point:

\( x=3, y=3 \)

Boundary Point:

\( x=0, y=0 \)

Boundary Point:

\( x=3, y=0 \)

Interior Critical Point:

\( z=0 \) at \( (1, 2) \)

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Example: Finding Extrema on a Triangular Region

Given the function:

\[ z = f(x,y) = x^2 + y^2 - 2x - 4y + 10 \]

Find the max/min above the triangle with vertices \((0,0)\), \((0,4)\), and \((4,0)\).

Paraboloid: \[ z = f(x,y) = (x-1)^2 + (y-2)^2 + 5 \]

Find the max/min of \(f(x,y)\) inside the triangle and on the boundary of the triangle.

Boundary Analysis

Notice the triangle can be broken down into 3 parts:

Figure: Right triangle on x-y axes with vertices (0,0), (4,0), and (0,4). Edges are numbered 1, 2, and 3.

On ①: \(0 \le x \le 4, y = 0\)
On ②: \(y = 4 - x, 0 \le x \le 4\)
On ③: \(0 \le y \le 4, x = 0\)

Strategy:

Find critical pts inside triangle
Then locations of max/min on each edge
Then compare

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Step 1: Find Interior Critical Points

Set partial derivatives to zero: \(f_x = 0, f_y = 0\)

\[ \begin{cases} f_x = 2x - 2 = 0 \\ f_y = 2y - 4 = 0 \end{cases} \]

CP: \((1, 2)\)

Is this inside the triangle? Yes, so we keep it.

Step 2: Analyze Boundary ①

Now we look at ①: \(y = 0, 0 \le x \le 4\)

\(f(x,y) = x^2 + y^2 - 2x - 4y + 10\) becomes:

\[ f(x) = x^2 - 2x + 10, \quad 0 \le x \le 4 \]

Notice this is reduced to a one-variable absolute max/min problem.

Critical pts: \(f'(x) = 2x - 2 = 0 \rightarrow x = 1, y = 0 \rightarrow\) \((1, 0)\)

Ends:

\(x = 0, y = 0 \rightarrow\) \((0, 0)\)
\(x = 4, y = 0 \rightarrow\) \((4, 0)\)

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Step 3: Analyze Boundary ②

On ②: \(y = 4 - x, 0 \le x \le 4\)

\(f(x,y) = x^2 + y^2 - 2x - 4y + 10\) becomes:

\[ f(x) = 2x^2 - 6x + 10, \quad 0 \le x \le 4 \]

Again, a one-variable absolute max/min.

\(f'(x) = 4x - 6 = 0 \rightarrow x = 3/2, y = 4 - x = 4 - 3/2 = 5/2\)

\((3/2, 5/2)\)

Ends:

\(x = 0, y = 4 - x \rightarrow\) \((0, 4)\)
\(x = 4, y = 4 - x \rightarrow\) \((4, 0)\)

Step 4: Analyze Boundary ③

On ③: \(x = 0, 0 \le y \le 4\)

\(f(x,y) = x^2 + y^2 - 2x - 4y + 10\) becomes:

\[ f(y) = y^2 - 4y + 10, \quad 0 \le y \le 4 \]

Following the same procedure as on ① and ② we get the following points of interest:

\((0, 2), (0, 0), (0, 4)\)

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Now we compare \( f(x,y) = x^2 + y^2 - 2x - 4y + 10 \) at each of those points of interest:

\( f(1,2) = 5 \) → abs. min at \( x=1, y=2, z=5 \) (vertex)
\( f(0,0) = 10 \)
\( f(4,0) = 18 \) → abs. max at \( x=4, y=0, f(x,y) = 18 \)
\( f(0,4) = 10 \)
\( f(3/2, 5/2) = 11/2 \)
\( f(0,2) = 6 \)
\( f(1,0) = 9 \)

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Contour Plot of \( z = f(x,y) = x^2 + y^2 - 2x - 4y + 10 \)

Figure: Contour plot of z=f(x,y) with concentric circles and a triangle with vertices (0,0), (4,0), and (0,4).

The plot displays level curves for \( z=5, z=6, z=8, \) and \( z=10 \). A triangle with vertices at \( (0,0), (4,0), \) and \( (0,4) \) is shown, representing the domain of interest.

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Example

Find abs max/min of \( f(x,y) = x^2 + y^2 - 6x + 9 \) → paraboloid above the region \( x^2 + y^2 \le 25 \) → circle.

Find highest/lowest on the paraboloid above the circle.

Figure: 3D diagram showing a paraboloid and its projection onto a circular region in the xy-plane.

Figure: 2D plot of a circle with radius 5 centered at the origin in the xy-plane.

\( x^2 + y^2 = 25 \)

\( y = \sqrt{25 - x^2} \) (upper)

\( y = -\sqrt{25 - x^2} \) (lower)

\[ f(x,y) = x^2 + y^2 - 6x + 9 \]\[ f_x = 2x - 6 = 0 \]\[ f_y = 2y = 0 \]

cp: (3, 0)

is this inside the restricted area? yes, so we keep it

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Contour Plot Analysis

The following graph shows the level curves for the function:

\[ t = f(x,y) = x^2 + y^2 - 6x + 9 \]

The plot displays concentric circles representing different values of \( t \), specifically labeled for \( t=10 \), \( t=20 \), and \( t=30 \). A large black circle centered at the origin with radius 5 is also shown, representing the boundary of the domain.

Figure: Contour plot of f(x,y) with level curves t=10, 20, 30 and a boundary circle of radius 5.

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Now let's travel along \( y = \sqrt{25 - x^2} \) and \( y = -\sqrt{25 - x^2} \), for \( -5 \le x \le 5 \). This corresponds to the boundary curve:

\[ x^2 + y^2 = 25 \]

Substituting this into our function:

\[ f(x,y) = x^2 + y^2 - 6x + 9 \]

Since \( x^2 + y^2 = 25 \), we have:

\[ f(x) = 25 - 6x + 9 = 34 - 6x \quad \text{for } -5 \le x \le 5 \]

Taking the derivative:

\[ f' = -6 \]

There are no critical points. Checking the endpoints:

At \( x = -5, y = 0 \)
At \( x = 5, y = 0 \)

(-5, 0)

(5, 0)

Comparison of Values

\( f(3, 0) = 0 \) — abs. min
\( f(-5, 0) = 64 \) — abs. max
\( f(5, 0) = 4 \)

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Example: Optimization on a Plane

What point on the plane \( 3x + 2y + z = 6 \) is closest to the origin?

Figure: 3D coordinate system showing a plane intersecting axes at (2,0,0), (0,3,0), and (0,0,6).

To find the point closest to the origin, we must minimize the distance \( d \):

\[ d = \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^2} = \sqrt{x^2 + y^2 + z^2} \]

We minimize this subject to the constraint:

\[ 3x + 2y + z = 6 \quad \text{or} \quad z = 6 - 3x - 2y \]

Substituting \( z \) into the distance formula:

\[ d = \sqrt{x^2 + y^2 + (6 - 3x - 2y)^2} \]

Since distance \( d \ge 0 \), minimizing \( d \) is equivalent to minimizing \( d^2 \). Let:

\[ f(x,y) = d^2 = x^2 + y^2 + (6 - 3x - 2y)^2 \]

To find the critical points, set the partial derivatives to zero: \( f_x = 0 \) and \( f_y = 0 \), then use the Second Derivative Test.