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23. Maximum and Minimum Values of Functions

Dated: 30-10-2024

Absolute Extrema

These are either the highest peaks in the mountain range or the deepest of valleys.

Absolute Minimum

If \(f(x_0) \le f(x)\) for all \(x\) in domain of \(f(x)\) then \(f(x_0)\) is called absolute minimum of \(f(x)\).

Absolute Maximum

If \(f(x_0) \ge f(x)\) for all \(x\) in domain of \(f(x)\) then \(f(x_0)\) is called absolute maximum of \(f(x)\).

Example

Pasted image 20240923082008.png

This is graph of function1 defined to be \(f(x) = 2x + 1\) over an interval2 \([0, 3)\).

\(x = 3\) is excluded from the domain, the maximum value for \(x\) is \(x < 3\).

The absolute minima for this function1 is \(1\).

You might think the absolute maxima for this function1 is \(7\) but it is actually not. Reason behind it is that the function1 never becomes equal to \(7\) since \(x\) never becomes \(3\), though, they are very close.

Theorem

If a \(f(x)\) has an extreme value (either maximum or minimum) on an open interval2 \((a, b)\) then these extreme values appear at critical points3 of \(f(x)\).

Finding Absolute Extrema for a Continuous Function

We can divide this task into following tasks.

  1. Find critical points3 of \(f(x)\).
  2. Find values of \(f(x)\) at critical points3 and the endpoints of the interval2 the function1 is defined at.
  3. The largest of the values in Step 2 is absolute maxima and the smallest is absolute minima.

Example

Find dimensions of a rectangle with \(100 ft\) perimeter such that it has maximum area.

Solution

We can represent the perimeter as:

\[100 = 2y + 2x\]
\[50 = x + y\]
\[y = 50 - x\]

Then we can represent the area which we want to maximize as:

\[A(x, y) = xy\]

We can re-write the function1 in terms of \(x\).

\[A(x) = x \cdot (50 - x) = 50x - x^2\]

From the equation above, it is evident that \(A(x) = 0\) if \(x = 0, 50\).

Since negative lengths are not realistic, \(x\) cannot go below \(0\) and if \(x > 50\) then the area becomes negative which is not realistic either. Hence, our interval2 is \([0, 50]\).

Then we find the absolute maxima. For that, we need to find the critical points.3 For that, we do our derivative test.4

\[A^{\prime}(x) = 50 - 2x\]
\[0 = 50 - 2x\]
\[2x = 50\]
\[x = 25\]

So our values to check are \(x = 0, 25, 50\).

\[A(0) = 50 \cdot 0 - 0^2 = 0\]
\[A(25) = 50 \cdot 25 - {25}^2 = 625\]
\[A(50) = 50 \cdot 50 - {50}^2 = 0\]

So the absolute maximum lies at \(x = 25\).
Put this in the perimeter equation.

\[y = 50 - x\]
\[y = 50 - 25 = 25\]

Hence, the dimensions of the rectangle with maximum area are \(25 \times 25\).

References

Read more about notations and symbols.

  1. Read more about functions. ↩↩↩↩↩↩

  2. Read more about intervals ↩↩↩↩

  3. Read more about critical points ↩↩↩↩

  4. Read more about derivative tests ↩