ENGI 3424 Engineering Mathematics

Faculty of Engineering and Applied Science

Example 4.8.3 (Max/Min on Finite Domain)

Question

In a test frame modelled by the unit square
              0 < x < 1 ,     0 < y < 1 ,
the potential field is     V (x, y)   =   48xy – 32x³ – 24y²

Find the absolute minimum and absolute maximum values of the potential field on the unit square
and the locations of the points where these absolute extrema occur.

Solution

In general, there are four categories of critical points to consider:


• on the boundary of the domain;
• where   V (x, y)   is undefined;
• where   V (x, y)   is undefined; and
• where   V (x, y) = 0.

The potential function is a polynomial function of   x   and   y , which guarantees continuity and differentiability everywhere in the domain.   Therefore only the first and last types of critical points need to be investigated in this example.

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Boundaries

V (x, y)   =   48xy – 32x3 – 24y2

Left edge: V (0, y)   =   0 – 0 – 24y2 For   0 < y < 1 ,     –24 < V (0, y) < 0 Bottom edge: V (x, 0)   =   0 – 32x3 – 0 For   0 < x < 1 ,     –32 < V (x, 0) < 0 Right edge: V (1, y)   =   48y – 32 – 24y2   =   –8(3(y–1)2 + 1) which is a downward opening parabola with its maximum at y = 1 V (1, 1)   =   –8(0 + 1)   =   –8 V (1, 0)   =   –8(3 + 1)   =   –32 For   0 < y < 1 ,     –32 < V (1, y) < –8

Top edge:
V (x, 1)   =   48x – 32x³ – 24
The derivative of this function (with respect to x) is
V ' (x, 1)   =   48 – 96x²   =   48(1 – 2x²)
The cubic function therefore has two real, distinct critical points,
a minimum at
and a maximum at
Only the maximum lies inside the domain.
V (0, 1) = –24 ,   a local minimum for this edge.
  a local maximum for this edge.
V (1, 1) = 48 – 32 – 24 = –8 ,   a local minimum for this edge.
Therefore, for   0 < x < 1 ,     –24 < V (x, 1) < –1.37 .

Therefore, on the boundary, the potential function varies between a minimum value of –32 at the point (1, 0) and a maximum value of 0 at the point (0, 0)

V (x, y)   =   48xy – 32x3 – 24y2

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Interior

V (x, y)   =   48xy – 32x³ – 24y²


y = x
x – 2x² = 0     x(1 – 2x) = 0
(x, y) = (0, 0) or (0.5, 0.5)
However, (0, 0) is on the boundary.
Therefore the only critical point in the interior is (0.5, 0.5).


This is greater than any value on the boundary.
It must therefore be the absolute maximum (for the unit square domain).

One may also use the second derivative test to verify that the point (0.5, 0.5, 2) is a relative maximum:


[Also note that if the domain were all real (x, y), then the only other critical point would be a saddle point at the origin, and that there would be no absolute extrema at all.]

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Maple plot of   z   =   48xy – 32x³ – 24y²,
viewed from   θ = 75°, φ = 300°:


Answer:

–32 < V (x, y) < +2   with the absolute maximum at (0.5, 0.5)   and the absolute minimum at (1, 0)

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