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PAGE 1

15.7 Max and Min Problems (part 1)

Recall if \( y = f(x) \), if \( f'(c) = 0 \) \( \rightarrow \)

  • \( c \) is a critical number
  • \( (c, f(c)) \) is a critical point
  • \( x = c \) is a possible location of relative max/min of \( f(x) \)

Then the Second Derivative Test says:

If \( f''(c) > 0 \rightarrow \) rel. min at \( x = c \)

concave up, so like \( \cup \)

If \( f''(c) < 0 \rightarrow \) rel. max at \( x = c \)

concave down, so like \( \cap \)

If \( f''(c) = 0 \rightarrow \) inconclusive

For \( z = f(x, y) \), a lot of this carry over, but some things change

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The two-variable version of Second Derivative Test:

Find critical pts: \( f_x = 0 \) AND \( f_y = 0 \)

collect critical pts \( (a, b) \)

Then we evaluate the discriminant

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

If at \( (a, b) \):

\( f_{xx} > 0 \) and \( D > 0 \rightarrow \) rel. min at \( (a, b) \)

\( f_{xx} < 0 \) and \( D > 0 \rightarrow \) rel. max at \( (a, b) \)

\( D < 0 \rightarrow \) saddle point at \( (a, b) \)

(neither max nor min)

\( D = 0 \rightarrow \) inconclusive

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\( D \) contains shape information near a critical point

For example, \( f(x, y) = x^2 + y^2 \rightarrow f_{xx} = 2, f_{yy} = 2, f_{xy} = 0 \)

3D graph of a paraboloid opening upwards from the origin (0,0).

at \( (0, 0) \) \( D = f_{xx} f_{yy} - (f_{xy})^2 > 0 \)

(rel. min)

  • concave up \( f_{xx} > 0 \)
  • concave up \( f_{yy} > 0 \)

Another example, \( f(x, y) = 4 - x^2 - y^2 \rightarrow f_{xx} = -2, f_{yy} = -2, f_{xy} = 0 \)

3D graph of a paraboloid opening downwards from a peak on the z-axis.

at \( (0, 0) \) \( D > 0 \)

Last example \( f(x, y) = -x^2 + y^2 \rightarrow f_{xx} = -2, f_{yy} = 2, f_{xy} = 0 \)

3D graph of a saddle-shaped surface centered at the origin.

at \( (0, 0) \) \( D < 0 \)

  • concave down \( f_{xx} = -2 \)
  • concave up \( f_{yy} = 2 \)

\( D > 0 \rightarrow \) near crit. pt parabolic approx

\( D < 0 \rightarrow \) can't do that

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Example

\[ f(x, y) = x^3 - 48xy + 64y^3 \]

Find critical pts: \( f_x = 0 \) AND \( f_y = 0 \)

\( f_x = 3x^2 - 48y + 0 = 0 \) — ①

\( f_y = -48x + 192y^2 = 0 \) — ②

from â‘  \( x^2 = 16y \)

from â‘¡ \( x = 4y^2 \)

\( (4y^2)^2 = 16y \)

\( 16y^4 - 16y = 0 \)

\( 16y(y^3 - 1) = 0 \)

\( y = 0 \) or \( y = 1 \)

these are the y-coord of critical pts

Then from \( x = 4y^2 \), we get the following crit. pts:

\[ (0, 0), (4, 1) \]
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\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]

\( f_{xx} = 6x \)

\( f_{yy} = 384y \)

\( f_{xy} = f_{yx} = -48 \)

\[ D = 2304xy - 2304 \]

Evaluate at critical pts

D(0, 0) < 0

\( \rightarrow \) saddle pt at (0, 0)

D(4, 1) > 0

\( f_{xx}(4, 1) > 0 \)

rel. min at (4, 1)

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Multivariable Calculus: Finding Critical Points

Example

\[ f(x,y) = xye^{-x^2-y^2} \]

Find critical points: \( f_x = 0 \) AND \( f_y = 0 \)

\[ f_x = \dots = ye^{-x^2-y^2}(-2x^2+1) = 0 \quad \text{--- (1)} \]
\[ f_y = \dots = xe^{-x^2-y^2}(-2y^2+1) = 0 \quad \text{--- (2)} \]

Analysis of Equation (1)

From (1): \( (e^{-x^2-y^2})(y)(-2x^2+1) = 0 \)

  • The term \( e^{-x^2-y^2} \) is an exponential and is never 0.
  • The terms \( (y)(-2x^2+1) \) are where critical points come from.

So,

\( y = 0 \)
or
\( x = \pm \frac{1}{\sqrt{2}} \)

do NOT pair these to form points because these came out of (1) = 0 so forming points with these will NOT guarantee (2) = 0

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Analysis of Equation (2)

From (2): \( (e^{-x^2-y^2})(x)(2y^2-1) = 0 \)

So,

\( x = 0 \)
or
\( y = \pm \frac{1}{\sqrt{2}} \)

Forming Critical Points

Pick one black boxed value from each to form critical points:

\[ (0,0), \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \]

Second Derivative Test

\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]
\[ f_{xx} = 2xy(2x^2-3)e^{-x^2-y^2} \]\[ f_{yy} = 2xy(2y^2-3)e^{-x^2-y^2} \]
\[ f_{xy} = (2x^2-1)(2y^2-1)e^{-x^2-y^2} \]
  • At \( (0,0) \), \( D < 0 \) → Saddle Point
  • At \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \), \( D > 0, f_{xx} < 0 \) → rel. max
  • At \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \), \( D > 0, f_{xx} > 0 \) → rel. min
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Case Study: When \( D = 0 \)

Example

\[ f(x,y) = xy^2 \]

\( (0,0) \) is a critical point.

\[ f_x = y^2, \quad f_y = 2xy, \quad f_{xy} = 2y, \quad f_{yy} = 2x, \quad f_{xx} = 0 \]
\[ D = f_{xx}f_{yy} - (f_{xy})^2 = -4y^2 \]

At \( (0,0) \), \( D = 0 \). Inconclusive: critical point can still be max/min/saddle point.

Sign Analysis Near Origin

One way to tell is to see the signs of \( f(x,y) = xy^2 \) near \( (0,0) \).

Cartesian plane showing f < 0 for x < 0 and f > 0 for x > 0, with f = 0 at the origin.

As we move right, \( f \) increases.

As we move left, \( f \) decreases.

So \( (0,0) \) is NOT a max or min.