Find the Maxima and Minima of an Equation

Extrema of a function are the maximum and minimum points. Maxima and minima are the highest and lowest values of a function within a set of ranges. The largest value of the function in the full range is called the absolute maximum, and the smallest value is called the absolute minimum, while the least value is known as the absolute minima. Other peaks and minima of a function are known as local maxima and minima, and they are not the absolute maxima and minima of the function. Let’s take a closer look at local maxima and minima, absolute maxima and minima, and how to calculate the function’s maximum and minima.

Maxima and Minima Points:

The place in a given interval where the values of the function near that point are always less than the value of the function at that point is known as a local maxima. Local minima, on the other hand, are points where the function’s values around that point are bigger than the function’s value at that point.

Maxima and Minima properties:

If f(x) is a continuous function in its domain, then there should be at least one maximum and one minimum between equal values of f(x) .

The maxima and minima alternate. In other words, there is one minimum between two maxima and vice versa.

If f(x) approaches infinity as x approaches an or b, and f'(x) = 0 for exactly one value x, namely c between a and b, then f(c) is the lowest and least value. If f(x) tends to – as x tends to an or b, f(c) is the greatest and most extreme value.

Maxima and Minima Examples:

Example 1: Find the points of maxima and minima of a function:

y = 2x³ – 3x² + 6

Solution

Given function: y = 2x³ – 3x² + 6

Using the second order derivative test to find a function’s maximum and minimum:

Taking the first derivative of:

y = 2x³ – 3x² + 6 —– (eq 1)

Differentiate both sides (eq 1), w.r.t x.

⇒ dy/dx = d(2x³)/dx – d(3x²)/dx + d(6)/dx

⇒ dy/dx = 6x² – 6x + 0

⇒ dy / dx = 6x² – 6x …. (eq 2)

To find crucial points, put dy/dx = 0.

⇒ 6x² – 6x = 0

⇒ 6x (x – 1) = 0

⇒ x = 0,1

The critical points are 0 & 1.

Differentiating both sides in eqn 2 w.r.t x.

⇒ d²y/dx² = d(6x²)/dx – d(6x)/dx

⇒ d²y/dx² = 12x – 6

Put the x values together and obtain the maximum or minimum value.

At x = 0, d²y/dx² = 12(0) – 6 = -6 < 0, hence x = 0 is a point of maxima

At x = 1, d²y/dx² = 12(1) – 6 = 6 > 0, hence x = 1 is a point of minima

The function’s maximum value is x = 0 and its minimum value is x = 1.

Example 2: Using the maxima and minima formulas, find the extrema and extremum value of the preceding function: f(x) = -3x² + 4x + 7.

Solution:

Using the second order derivative test to find a function’s maximum and minimum:

Given function: f(x) = -3x² + 4x + 7 —————- (eq 1)

On both sides of (eq 1), differentiate w.r.t x.

⇒ dy/dx = d(-3x²)/dx + d(4x)/dx + d(7)/dx

⇒ dy/dx = – 6x + 4

dy/dx = 0 helps find critical points.

⇒ -6x + 4 = 0 ——-(eq 2)

⇒x = 2/3

The critical point is 2/3.

w.r.t. x, differentiate both sides of (eq 2)

⇒ d²y/dx² = d(-6x)/dx + d(4)/dx

⇒ d²y/dx² = -6

Since d²y/dx² < 0, the given curve will have maxima at x = 2/3.

At x = 2/3, the maximum value of f(x) is

f(2/3) = -3(2/3)

2 + 4(2/3) + 7 = -4/3 + 8/3 + 7 = 25/3

The maxima of the function is at x = 2/3 and maximum value is 25/3.

Conclusion:

There are many times when you need to know the highest or lowest possible value of a quantity. Such applications can be found in economics, business, and engineering, to name a few. Many of them can be solved using the differential calculus methods discussed above.