MA 261 — Lesson 15: Max and Min Problems I (Section 15.7)

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Page 1: Course announcements including Exam 1 date, location, time, and office hours schedule
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Announcements

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

The following items are posted on Brightspace: instructions, seating chart, and study guide.

Office Hours for Exam 1:

Date(s)Time
02/16, 02/18, 02/209:45 – 11:15
02/239:45 – 11:45
02/24, 02/252:00 pm – 3:00 pm

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Page 2: Warmup example reviewing critical points and second derivative test for single-variable function f(x) = x squared (x minus 1)(x minus 2)
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Warmup Example: Single-Variable Review

Consider \(f(x) = x^2(x-1)(x-2)\).

Step 1 — Find Critical Points

A value \(x = a\) is a critical point if \(f'(a) = 0\) or \(f'(a)\) does not exist (DNE).

Setting \(f'(x) = 0\) and solving:

\[f'(x) = 0 \implies x = 0,\; x = 1,\; x = 2 \quad \text{(3 critical points)}\]
Step 2 — Classify via the Second Derivative Test

Second Derivative Test (one variable): For a critical point \(x = a\),

  • \(f''(a) > 0 \implies\) local minimum
  • \(f''(a) < 0 \implies\) local maximum
  • \(f''(a) = 0 \implies\) inconclusive

Solution: Using the product rule,

\[f'(x) = 2x(x-1)(x-2) + x^2(x-2) + x^2(x-1)\]

Evaluating \(f''\) at each critical point:

  • \(f''(0) = 0\) → inconclusive
  • \(f''(1) = -1 < 0\) → \(x = 1\) is a local maximum
  • \(f''(2) = 4 > 0\) → \(x = 2\) is a local minimum

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Page 3: Definitions of critical points and second derivative test for two-variable functions z = f(x,y) with the discriminant D
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Two-Variable Functions: \(z = f(x,y)\)

Critical Point: A point \((a,b)\) is a critical point of \(f(x,y)\) if

\[\nabla f(a,b) = \langle 0,0\rangle \quad \text{i.e.,} \quad f_x(a,b) = 0 \;\text{ and }\; f_y(a,b) = 0\]

or one of the partial derivatives does not exist.

Second Derivative Test (two variables): Let \((a,b)\) be a critical point. Define the discriminant

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

Then:

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Page 4: Three geometric examples illustrating local minimum bowl shape, local maximum hill shape, and saddle point with their discriminant calculations
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Geometric Examples of the Second Derivative Test

Example 1: \(z = x^2 + y^2\) — Local Minimum (Bowl Shape)

Diagram description: A bowl-shaped paraboloid opening upward, with a highlighted point at the bottom of the bowl labeled "local min." The surface curves upward in all directions from the lowest point, similar to the inside of a cereal bowl.

\(f_x = 2x = 0\) and \(f_y = 2y = 0\) gives critical point \((0,0)\).

\(f_{xx} = 2 > 0\), \(f_{yy} = 2 > 0\), \(f_{xy} = 0\), so \(D = 4 > 0\).  → Local minimum.

Example 2: \(z = -(x^2 + y^2)\) — Local Maximum (Hill Shape)

Diagram description: A hill-shaped paraboloid opening downward, with the peak highlighted at the top labeled "Max." The surface curves downward in all directions from the highest point.

Critical point \((0,0)\). \(f_{xx} = f_{yy} = -2 < 0\), \(f_{xy} = 0\), so \(D = 4 > 0\).  → Local maximum.

Example 3: \(z = x^2 - y^2\) — Saddle Point

Diagram description: A saddle-shaped surface centered at the origin. A red curve through the saddle point curves upward (like a local minimum in the \(x\)-direction) while blue/green curves curve downward (like a local maximum in the \(y\)-direction). The surface resembles a mountain pass or horse saddle. The critical point in the center is neither a max nor a min.

Critical point \((0,0)\). \(f_{xx} = 2\), \(f_{yy} = -2\), \(f_{xy} = 0\), so \(D = -4 < 0\).  → Saddle point.

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Page 5: Finding three critical points of f(x,y) = x to the fourth plus y to the fourth minus 4xy plus 1 by substitution
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Example: \(f(x,y) = x^4 + y^4 - 4xy + 1\)

Finding Critical Points

Setting partial derivatives to zero:

\[f_x = 4x^3 - 4y = 0 \implies y = x^3 \tag{1}\] \[f_y = 4y^3 - 4x = 0 \implies y^3 = x \tag{2}\]

Substitute (1) into (2): \((x^3)^3 = x \implies x^9 = x \implies x(x^8 - 1) = 0\), giving \(x = 0\), \(x = 1\), \(x = -1\).

Substituting back into (1):

  • \(x = 0 \implies y = 0\)
  • \(x = 1 \implies y = 1\)
  • \(x = -1 \implies y = -1\)

There are 3 critical points: \((0,0)\), \((1,1)\), and \((-1,-1)\).

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Page 6: Second derivative test table for f(x,y) = x to the fourth plus y to the fourth minus 4xy plus 1, showing saddle point at origin and local minima at (1,1) and (-1,-1)
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Second Derivative Test for \(f(x,y) = x^4 + y^4 - 4xy + 1\)

Second-order partial derivatives:

\[f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4\] \[D = f_{xx}f_{yy} - (f_{xy})^2 = 144x^2y^2 - 16\]
Point\(f_{xx}\)\(f_{yy}\)\(f_{xy}\)\(D\)Classification
\((0,0)\)\(0\)\(0\)\(-4\)\(-16 < 0\)Saddle Point
\((1,1)\)\(12 > 0\)\(12\)\(-4\)\(144-16 > 0\)Local Min
\((-1,-1)\)\(12 > 0\)\(12\)\(-4\)\(144-16 > 0\)Local Min

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Page 7: Finding five critical points of f(x,y) = x y e to the negative (2x squared plus 8y squared) using the product rule and solving two simultaneous equations
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Example: \(f(x,y) = xy\,e^{-(2x^2+8y^2)}\)

Finding Critical Points

Since \(e^{-(2x^2+8y^2)} > 0\) always, it factors out of both partial derivatives:

\[f_x = e^{-(2x^2+8y^2)}\cdot y(1-4x^2) = 0 \implies y(1-4x^2) = 0 \tag{1}\] \[f_y = e^{-(2x^2+8y^2)}\cdot x(1-16y^2) = 0 \implies x(1-16y^2) = 0 \tag{2}\]

From (1): \(y = 0\) or \(x = \pm\tfrac{1}{2}\).

Case \(y = 0\): Equation (2) becomes \(x(1) = 0 \implies x = 0\). Critical point: \((0,0)\).

Case \(x = \tfrac{1}{2}\): Equation (2) gives \(\tfrac{1}{2}(1-16y^2)=0 \implies y = \pm\tfrac{1}{4}\). Critical points: \(\bigl(\tfrac{1}{2},\tfrac{1}{4}\bigr)\) and \(\bigl(\tfrac{1}{2},-\tfrac{1}{4}\bigr)\).

Case \(x = -\tfrac{1}{2}\): Similarly \(y = \pm\tfrac{1}{4}\). Critical points: \(\bigl(-\tfrac{1}{2},\tfrac{1}{4}\bigr)\) and \(\bigl(-\tfrac{1}{2},-\tfrac{1}{4}\bigr)\).

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Page 8: Second derivative test table for f(x,y) = xy e to the negative (2x squared plus 8y squared), showing one saddle point at origin, two local maxima, and two local minima
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Second Derivative Test for \(f(x,y) = xy\,e^{-(2x^2+8y^2)}\)

Second-order partial derivatives (the exponential factor is always positive):

\[f_{xx} = -4xy(3-4x^2)\,e^{-(2x^2+8y^2)}\] \[f_{yy} = -16xy(3-16y^2)\,e^{-(2x^2+8y^2)}\] \[f_{xy} = (1-4x^2)(1-16y^2)\,e^{-(2x^2+8y^2)}\]

Key observation: Since \(e^{-(2x^2+8y^2)} > 0\) always, the sign of each second-order partial is determined entirely by its polynomial prefactor. Let \(k = e^{-1} > 0\) denote the exponential value at the non-origin critical points.

Point\(f_{xx}\)\(f_{yy}\)\(f_{xy}\)\(D\)Classification
\((0,0)\)\(0\)\(0\)\(1\)\(-1 < 0\)Saddle Point
\(\bigl(\tfrac{1}{2},\tfrac{1}{4}\bigr)\) \(-\tfrac{1}{2}{\cdot}2{\cdot}k < 0\) \(-2{\cdot}2{\cdot}k < 0\) \(0\)\(> 0\)Local Max
\(\bigl(\tfrac{1}{2},-\tfrac{1}{4}\bigr)\) \(\tfrac{1}{2}{\cdot}2{\cdot}k > 0\) \(2{\cdot}2{\cdot}k > 0\) \(0\)\(> 0\)Local Min
\(\bigl(-\tfrac{1}{2},\tfrac{1}{4}\bigr)\) \(\tfrac{1}{2}{\cdot}2{\cdot}k > 0\) \(2{\cdot}2{\cdot}k > 0\) \(0\)\(> 0\)Local Min
\(\bigl(-\tfrac{1}{2},-\tfrac{1}{4}\bigr)\) \(-\tfrac{1}{2}{\cdot}2{\cdot}k < 0\) \(-2{\cdot}2{\cdot}k < 0\) \(0\)\(> 0\)Local Max

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Page 9: Finding two critical points of f(x,y) = one-third y cubed plus x squared plus 4xy minus 2x minus 13y plus 7 by substitution
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Example: \(f(x,y) = \tfrac{1}{3}y^3 + x^2 + 4xy - 2x - 13y + 7\)

Finding Critical Points

Setting partial derivatives to zero:

\[f_x = 2x + 4y - 2 = 0 \tag{1}\] \[f_y = y^2 + 4x - 13 = 0 \tag{2}\]

From (1): \(x = 1 - 2y\). Substituting into (2):

\[y^2 + 4(1-2y) - 13 = 0 \implies y^2 - 8y - 9 = 0 \implies (y-9)(y+1) = 0\] \[\implies y = 9 \quad \text{or} \quad y = -1\]

Back-substituting into \(x = 1 - 2y\):

  • \(y = 9 \implies x = 1 - 18 = -17\). Critical point: \((-17,\, 9)\).
  • \(y = -1 \implies x = 1 + 2 = 3\). Critical point: \((3,\, -1)\).

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Page 10: Second derivative test for f(x,y) = one-third y cubed plus x squared plus 4xy minus 2x minus 13y plus 7, showing local minimum at (-17,9) and saddle point at (3,-1)
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Second Derivative Test for \(f(x,y) = \tfrac{1}{3}y^3 + x^2 + 4xy - 2x - 13y + 7\)

Second-order partial derivatives:

\[f_{xx} = 2, \quad f_{yy} = 2y, \quad f_{xy} = 4\]
Point\(f_{xx}\)\(f_{yy}\)\(f_{xy}\)\(D = f_{xx}f_{yy}-(f_{xy})^2\)Classification
\((-17,9)\)\(2\)\(18\)\(4\) \(36 - 16 = 20 > 0\)Local Min
\((3,-1)\)\(2\)\(-2\)\(4\) \(-4 - 16 = -20 < 0\)Saddle Point

Summary: At \((-17,9)\), \(D > 0\) and \(f_{xx} = 2 > 0\), confirming a local minimum. At \((3,-1)\), \(D < 0\), confirming a saddle point.