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Dated: 30-10-2024

Examples

Example

Find absolute maximum1 and absolute minimum1 values on the closed triangular region \(R\) with vertices \((0, 0)\), \((0, 4)\) and \((5, 0)\) for the function2

\[f(x, y) = xy - x - 3y\]

Solution

\[f_x(x, y) = y - 1\]
\[f_y(x, y) = x - 3\]

For Critical Points

\[f_x(x, y) = 0\text{ and } f_y(x, y) = 0\]
\[y = 1\]
\[x = 3\]

So there is only one critical point3 that is \((3, 1)\).

Boundaries

There are 3 boundaries for triangle which are line segments.4

Line between \((0, 0)\) and \((5, 0)\)

Here \(y = 0\), therefore,

\[f(x, 0) = -x \text{ where } 0 \le x \le 5\]
Line between \((0, 0)\) and \((0, 4)\)

Here \(x = 0\), therefore,

\[f(0, y) = -3y \text{ where } 0 \le y \le 4\]
Line between \((5, 0)\) and \((0, 4)\)

Here \(y = - \frac 4 5 x + 4\), therefore,

\[f\left(x, -\frac 4 5 x + 4\right) = - \frac 4 5 x^2 + \frac 3 5 x + 12 \text{ where } 0 \le x \le 5\]
\[f^{\prime}\left(x, -\frac 4 5 x + 4\right) = - \frac 8 5 x + \frac 3 4 x\]

If we solve for \(x\), we get \(x = \frac 3 8\).
Similarly, changing the equation into

\[-(y - 4) \cdot \frac 5 4 = x\]

putting it inside \(f(x, y)\) and solving for \(y\), we get

\[y = \frac{37}{10}\]
\((x, y)\) \((f(x, y))\)
\((0, 0)\) \(0\)
\((5, 0)\) \(-5\)
\((0, 4)\) \(-4\)
\(\left(\frac{3}{8}, \frac{37}{10}\right)\) \(- \frac{807}{80}\)
\((3, 1)\) \(-3\)

From there, we can conclude that absolute maximum1 is \(0\) existing at \((0, 0)\) and absolute minimum1 is \(- \frac{807}{80}\) existing at \((\frac{3}{8}, \frac{37}{10})\).

Example

Find dimensions of a box such that is has maximum volume, which is inscribed in a sphere with radius \(r = 4\).

Solution

The equation for volume is

\[V(x, y, z) = xyz\]

The equation for sphere is

\[x^2 + y^2 + z^2 = 4^2\]

We will isolate \(z\),

\[z = \sqrt{16 - x^2 - y^2}\]

Plugging it back into \(V(x, y, z)\), the function2 becomes dependent on only \(x\) and \(y\).

\[V(x, y) = xy \sqrt{16 - x^2 - y^2}\]

Since we want to maximize it, the absolute maximum3 exists at critical points.3
To find those, we assume \(V_x = 0\) and \(V_y = 0\).

\[V_x = \frac{\partial}{\partial x} xy\sqrt{16 - x^2 - y^2}\]
\[0 = y\left(\frac{-2x^{2}-y^{2}+16}{\sqrt{16-x^{2}-y^{2}}}\right)\]

This gets us down to

\[2x^2 + y^2 = 16\]

Similarly, \(V_y\) gets us down to

\[x^2 + 2y^2 = 16\]

Solving both these equations, we get

\[x = y = \frac{4}{\sqrt{3}}\]

This suggests there is only one critical point.3
Then for our 2nd partial derivative test,5 we find additional terms which are

\[V_{xx}=\frac{xy(2x^{2}+3y^{2}-48)}{(16-x^{2}-y^{2})^{\frac{3}{2}}}\]
\[V_{yy}=\frac{xy(3x^{2}+2y^{2}-48)}{(16-x^{2}-y^{2})^{\frac{3}{2}}}\]
\[V_{xy} = \frac{2y^{4}+3x^{2}y^{2}-48y^{2}-2x^{2}y-16x^{2}+256}{(16-x^{2}-y^{2})\sqrt{16-x^{2}-y^{2}}}\]
\[\because D = V_{xx}(x_0, y_0) \cdot V_{yy}(x_0, y_0) - V_{xy}^2(x_0, y_0)\]
\[V_{xx}\left(\frac{4}{\sqrt 3}, \frac{4}{\sqrt 3}\right) = \frac{-16}{3}\]
\[V_{yy}\left(\frac{4}{\sqrt 3}, \frac{4}{\sqrt 3}\right) = \frac{-16}{3}\]
\[V_{xy}\left(\frac{4}{\sqrt 3}, \frac{4}{\sqrt 3}\right) = \frac{-8}{3}\]

Plugging it into \(D\), we get

\[D = \frac {320}{3} > 0\]

Since \(D > 0\) and \(V_{xx} < 0\), the absolute maximum3 exists at \(\left(\frac{4}{\sqrt 3}, \frac{4}{\sqrt 3}\right)\).
Therefore, the side lengths are \(\frac 4 {\sqrt 3}\).

References

Read more about notations and symbols.

  1. Read more about extreme values. ↩↩↩↩

  2. Read more about functions. ↩↩

  3. Read more about extreme values. ↩↩↩↩↩

  4. Read more about lines. ↩

  5. Read more about second partial derivative test. ↩