MA 261 - Lesson 16: Max and Min Problems II (Section 15.7)

Announcements

Exam 1 is on February 25th (Wednesday) in ELTT 116 at 6:30 pm.

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Review Example: Applying the Second Derivative Test

Review Question

Suppose \(f_{xx}(a,b) \neq 0\), \(\nabla f(a,b) = \vec{0}\) (so \((a,b)\) is a critical point), and \(f_{xx}(a,b) + f_{yy}(a,b) = 0\). What can you say about the point \((a,b)\)?

Options: A) Saddle B) Local Max C) Local Min D) Cannot be Determined E) I don't know

Answer: A) Saddle.

The condition \(f_{xx} + f_{yy} = 0\) means \(f_{xx} = -f_{yy}\). Substituting into the discriminant:

\[D = f_{xx}f_{yy} - (f_{xy})^2 = -(f_{yy})^2 - (f_{xy})^2 \leq 0\]

Since \(f_{xx} \neq 0\), we have \(f_{yy} \neq 0\), so \(D = -(f_{yy})^2 - (f_{xy})^2 < 0\). Therefore \((a,b)\) is a saddle point.

Reminder — Second Derivative Test summary:

\(D > 0\) and \(f_{xx} > 0\): local min
\(D > 0\) and \(f_{xx} < 0\): local max
\(D < 0\): saddle point
\(D = 0\): inconclusive

Absolute Max/Min for \(y = f(x)\) on \([a,b]\)

Graph description: A continuous curve on the interval \([-4, 5]\) showing multiple local extrema. The graph rises steeply from a minimum near \((-4, -4)\) to a local maximum near \((-2, 5)\), then oscillates through additional local extrema. Two highlighted points (shown as filled dots) appear at the global minimum around \((-4, -4)\) and a boundary point near \((5, 2.5)\). A horizontal dashed red line at \(y = 5\) marks the global maximum level.

The absolute maximum and minimum of \(f(x)\) on a closed interval \([a,b]\) occur either at critical points inside \((a,b)\) or at the endpoints/boundary points.

Process for Finding Absolute Extrema on \([a,b]\):

Find all critical points of \(f(x)\) inside \((a,b)\).
Evaluate \(f(x)\) at the critical points and at the endpoints \(x = a\) and \(x = b\).
Compare all values — the largest is the absolute maximum and the smallest is the absolute minimum.

Example: \(g(x) = 2x^2 - 1\) on \([-1, 1]\)

Step 1 — Critical points: \(g'(x) = 4x = 0 \implies x = 0\).

Step 2 — Evaluate:

\(g(0) = -1\)
\(g(-1) = 2(1) - 1 = 1\)
\(g(1) = 2(1) - 1 = 1\)

Step 3 — Compare:

Absolute minimum \(= -1\) at \(x = 0\).
Absolute maximum \(= 1\) at \(x = 1\) and \(x = -1\).

Absolute Max/Min of \(z = f(x,y)\) Over a Region

Diagram description: A closed, bounded region in the \(xy\)-plane drawn as an irregular blob shape (similar to an amoeba or figure-eight). The interior is filled with blue diagonal lines (hatching) indicating the region, and the boundary is drawn as a purple curve. This represents a generic closed bounded region \(D\) over which we seek the absolute maximum and minimum of \(f\).

For a continuous function \(f(x,y)\) on a closed bounded region, the absolute maximum and minimum occur either at critical points inside the region or on the boundary of the region.

Process for Finding Absolute Extrema Over a Region:

Find all critical points inside the region (where \(\nabla f = \vec{0}\)).
Use the boundary to write \(f(x,y)\) as a function of one variable, then apply Calculus I techniques to find additional candidates along each boundary segment.
Evaluate \(f\) at all candidates and compare — the largest value is the absolute maximum and the smallest is the absolute minimum.

Example: \(f(x,y) = x^2 - y^2\) on \(x^2 + y^2 \leq 1\)

Diagram description: A unit circle \(x^2 + y^2 = 1\) drawn in purple, with the interior region \(x^2 + y^2 \leq 1\) filled with blue diagonal hatching. A red dot marks the origin \((0,0)\) as the only interior critical point. The labels indicate the boundary circle and the closed disk region.

Step 1 — Critical Points Inside the Disk

\(f_x = 2x = 0\) and \(f_y = -2y = 0\), so \((0,0)\) is the only critical point.

Step 2 — On the Boundary \(x^2 + y^2 = 1\)

On the boundary, \(y^2 = 1 - x^2\). Substitute into \(f\) to eliminate \(y\):

\[g(x) = x^2 - (1 - x^2) = 2x^2 - 1, \quad -1 \leq x \leq 1\]

Boundary Analysis (continued): \(g(x) = 2x^2 - 1\) on \([-1,1]\)

Diagram description: The unit circle with four red dots highlighted at the cardinal points: top \((0,1)\), bottom \((0,-1)\), left \((-1,0)\), and right \((1,0)\). These are the boundary candidate points identified from the analysis of \(g(x)\).

Critical point of \(g\): \(g'(x) = 4x = 0 \implies x = 0\). Substituting \(x = 0\) into the boundary equation \(x^2 + y^2 = 1\) gives \(y = \pm 1\), yielding boundary points \((0, 1)\) and \((0, -1)\).

Endpoints of \(g\):

\(x = -1\): \((-1)^2 + y^2 = 1 \implies y = 0\), giving point \((-1, 0)\).
\(x = 1\): \(1 + y^2 = 1 \implies y = 0\), giving point \((1, 0)\).

The four boundary candidate points are \((0,1)\), \((0,-1)\), \((-1,0)\), and \((1,0)\).

Evaluate & Compare for \(f(x,y) = x^2 - y^2\)

All possible candidates for the absolute max/min: \((0,0)\), \((1,0)\), \((-1,0)\), \((0,1)\), \((0,-1)\).

Point	\(f(x,y) = x^2 - y^2\)
\((0,0)\)	\(0\)
\((1,0)\)	\(1\)
\((-1,0)\)	\(1\)
\((0,1)\)	\(-1\)
\((0,-1)\)	\(-1\)

Absolute maximum \(= 1\) at \((1,0)\) and \((-1,0)\).

Absolute minimum \(= -1\) at \((0,1)\) and \((0,-1)\).

Example: \(f(x,y) = 4x + 6y - x^2 - y^2 + 9\) on \(\{(x,y): 0 \leq x \leq 4,\; 0 \leq y \leq 5\}\)

Diagram description: A rectangle in the \(xy\)-plane with corners at \((0,0)\), \((4,0)\), \((4,5)\), and \((0,5)\) drawn in purple. The interior is filled with blue hatching. A red dot at the interior critical point \((2,3)\) is visible. The four boundary segments are labeled: \(L_1\) (bottom, \(y=0\)), \(L_2\) (right, \(x=4\)), \(L_3\) (top, \(y=5\)), and \(L_4\) (left, \(x=0\)).

Step 1 — Critical Points Inside the Rectangle

\(f_x = 4 - 2x = 0 \implies x = 2\)

\(f_y = 6 - 2y = 0 \implies y = 3\)

Critical point: \((2, 3)\) — this lies inside the rectangle since \(0 < 2 < 4\) and \(0 < 3 < 5\). ✓

Step 2 — Identify Boundary Segments

\(L_1\): \(y = 0\), \(0 \leq x \leq 4\)
\(L_2\): \(x = 4\), \(0 \leq y \leq 5\)
\(L_3\): \(y = 5\), \(0 \leq x \leq 4\)
\(L_4\): \(x = 0\), \(0 \leq y \leq 5\)

Boundary Analysis: \(L_1\) — \(y = 0\), \(0 \leq x \leq 4\)

Substituting \(y = 0\) into \(f\):

\[g_1(x) = 4x - x^2 + 9, \quad 0 \leq x \leq 4\]

Critical point: \(g_1'(x) = 4 - 2x = 0 \implies x = 2\). This gives candidate point \((2, 0)\).

Endpoints:

\(x = 0 \implies (0, 0)\)
\(x = 4 \implies (4, 0)\)

Diagram description: The rectangle with boundary segments labeled \(L_1\) through \(L_4\). Three red dots are visible along the bottom edge (\(L_1\)) at \(x = 0\), \(x = 2\), and \(x = 4\), corresponding to the two endpoints and the critical point found along this segment.

Boundary Analysis: \(L_2\) — \(x = 4\), \(0 \leq y \leq 5\)

Substituting \(x = 4\) into \(f\):

\[g_2(y) = 16 + 6y - 16 - y^2 + 9 = 6y - y^2 + 9, \quad 0 \leq y \leq 5\]

Critical point: \(g_2'(y) = 6 - 2y = 0 \implies y = 3\). Candidate: \((4, 3)\).

Endpoints:

\(y = 0 \implies (4, 0)\)
\(y = 5 \implies (4, 5)\)

Boundary Analysis: \(L_3\) and \(L_4\)

\(L_3\): \(y = 5\), \(0 \leq x \leq 4\)

Substituting \(y = 5\):

\[g_3(x) = 4x + 30 - x^2 - 25 + 9 = 4x - x^2 + 14, \quad 0 \leq x \leq 4\]

Critical point: \(g_3'(x) = 4 - 2x = 0 \implies x = 2\). Candidate: \((2, 5)\).

Endpoints: \(x = 0 \implies (0,5)\); \(x = 4 \implies (4,5)\).

\(L_4\): \(x = 0\), \(0 \leq y \leq 5\)

Substituting \(x = 0\):

\[g_4(y) = 6y - y^2 + 9, \quad 0 \leq y \leq 5\]

Critical point: \(g_4'(y) = 6 - 2y = 0 \implies y = 3\). Candidate: \((0, 3)\).

Endpoints: \(y = 0 \implies (0,0)\); \(y = 5 \implies (0,5)\).

Evaluate & Compare for \(f(x,y) = 4x + 6y - x^2 - y^2 + 9\)

Diagram description: The rectangle showing all candidate points plotted as labeled red dots: interior point \((2,3)\), boundary points \((2,0)\), \((4,3)\), \((2,5)\), \((0,3)\), and corner points \((0,0)\), \((4,0)\), \((0,5)\), \((4,5)\).

Point	\(f(x,y)\)
\((0,0)\)	\(9\)
\((2,0)\)	\(13\)
\((4,0)\)	\(9\)
\((0,3)\)	\(18\)
\((2,3)\)	\(22\)
\((4,3)\)	\(18\)
\((0,5)\)	\(14\)
\((2,5)\)	\(18\)
\((4,5)\)	\(14\)

Absolute minimum \(= 9\) at \((0,0)\) and \((4,0)\).

Absolute maximum \(= 22\) at \((2,3)\).

Example: Closest Point on a Plane to the Origin

Find the point on the plane \(x + 2y + z = 10\) that is closest to the origin.

Setting Up the Optimization

The distance from the origin \((0,0,0)\) to a point \((x,y,z)\) is

\[d = \sqrt{x^2 + y^2 + z^2}\]

Since \((x,y,z)\) must lie on the plane \(x + 2y + z = 10\), we can express \(z = 10 - x - 2y\) and substitute:

\[d = \sqrt{x^2 + y^2 + (10-x-2y)^2}\]

Key trick: Minimizing \(d\) is equivalent to minimizing \(d^2\) (since the square root is increasing). This avoids the square root in our calculus computations.

Rephrase: Find the minimum of \(f(x,y) = x^2 + y^2 + (10-x-2y)^2\) on \(\mathbb{R}^2\).

Solving: Minimum of \(f(x,y) = x^2 + y^2 + (10-x-2y)^2\)

Setting partial derivatives to zero:

\[f_x = 2x + 2(10-x-2y)(-1) = 4x + 4y - 20 = 0 \tag{1}\] \[f_y = 2y + 2(10-x-2y)(-2) = 4x + 10y - 40 = 0 \tag{2}\]

Subtract (1) from (2): \(6y - 20 = 0 \implies y = \dfrac{20}{6} = \dfrac{10}{3}\).

Substitute \(y = \tfrac{10}{3}\) into (1): \(4x + \tfrac{40}{3} - 20 = 0 \implies 4x = \tfrac{20}{3} \implies x = \dfrac{5}{3}\).

One critical point: \(\left(\dfrac{5}{3},\, \dfrac{10}{3}\right)\).

Verification that this is a minimum:

\(f_{xx} = 4 > 0\)
\(f_{yy} = 10 > 0\)
\(f_{xy} = 4\)
\(D = (4)(10) - 4^2 = 40 - 16 = 24 > 0\)

Since \(D > 0\) and \(f_{xx} > 0\), the critical point is a local (and global) minimum.

Finding \(z\):

\[z = 10 - x - 2y = 10 - \frac{5}{3} - \frac{20}{3} = 10 - \frac{25}{3} = \frac{5}{3}\]

The point on the plane closest to the origin is

\[\left(\frac{5}{3},\; \frac{10}{3},\; \frac{5}{3}\right)\]