Maxima and Minima

Engineering Context

ECE:

We can use the method for finding maximum and minimums to find the maximum current of an inductor. For an inductor in a circuit, the derivative of the current is voltage.

Current is given as:

\[ i(t) = te^{(-3t)} \]

and the formula for voltage is:

\[ v(t) = L \frac{di(t)}{dt} \]

Where L is the given inductance of the inductor in henry (H). In this case, we will say \( L = 200 \, \text{mH} \). Thus, the voltage is derived from \( i(t) \).

\[ v(t) = 0.2e^{-3t}(-3t + 1) \]

(1)

The current is at its maximum when its slope, or v(t) equals zero. So we set the voltage equation equal to zero and solve for the t time at which the current is at its maximum.

\[ v(t) = 0.2e^{-3t}(-3t + 1) = 0 \]
\[ t = \frac{1}{3} \]

Now, we can plug this time into the formula for current to find the maximum current.

\[ i ({\frac{1}{3}}) = (\frac{1}{3})e^{-3\left(\frac{1}{3}\right)} = \frac{1}{3}e^{-1} \]
\[ i_\text{max} = 0.123 \]

We can visualize this maximum current on a graph.

graph of Electrical example

CEE:

Cantilever beams are used frequently amongst civil engineers to build structures, such as parking garages and balconies. To find the maximum displacement of a cantilever beam when a mass is impacting it, we can use the essentials of extrema. The position of a cantilever beam is given as:

\[ y(t) = \frac{v_0}{\omega} \sin(\omega t) \]

Where an object of mass m is moving at a velocity \( v_0 \), impacting a cantilever beam with a given length l and flexural rigidity EI. The angular frequency of displacement is given as:

\[ \omega = \sqrt{\frac{3EI}{ml^3}} \]

This equation, combined with the essentials of maxima and minima, allows us to find the maximum displacement of the beam. We do this by setting the velocity equal to zero, solving for the times t when the displacement is a possible maximum, testing the times t for a maximum with the second derivative test, and then plugging the resulting time t into the position equation. Velocity is the derivative of position:

\[ v(t) = \frac{dy(t)}{dt} \]

Thus,

\[ v(t) = v_0 \cos\omega t = 0 \]
\[ t = \frac{\pi}{2\omega}, \frac{3\pi}{2\omega} \]

Using the second derivative test, we find that \( t = \frac{\pi}{2\omega} \) gives us the maximum displacement. Thus,

\[y_\text{max} = \frac{v_0}{\omega} \sin\frac{\omega\pi}{2\omega} \]

Because we know that \( \omega = \sqrt{\frac{3EI}{ml^3}} \) and \( \sin\frac{\pi}{2} = 1 \) , the maximum displacement of the cantilever beam is given as:

\[ y_\text{max} = v_0\frac{\sqrt{ml^3}}{\sqrt{3EI}} \]

The Essentials

Stated simply, the maxima and minima are the largest and smallest values of a function, and are otherwise known as extrema. They can be visualized as the peak of a mountain or depth of a valley, which helps us create accurate graphs. Such extrema can be either absolute or local; an absolute extremum is the definite largest or smallest value of a function, as represented in Figures 1 and 2.

a) describes a function where the output (denoted as f(x)) is equal to the square of the input value x

(a) \( f(x) = x^2 \)

b) represents a function where the output (denoted as f(x)) is equal to the negative square of the input value x.

(b) \( f(x) = -x^2 \)

Whereas a local extremum is not the definite minimum or maximum of the function, but it is relatively higher or smaller than the surrounding points within a range.

a) describes a function where the output, denoted as f(x), is equal to one-third times the cube of the input value x, minus seven halves times the square of x, plus five times x.

(a) \( f(x) = \frac{1}{3}x^3 - \frac{7}{2}x^2 + 5x \)

b) describes a function where the output, denoted as f(x), is equal to x raised to the power of the operation "**4", minus 4 times x raised to the power of the operation "**2", minus 2

(b) \( f(x) = x**4 - 4 * x**2 - 2 \)

We can find the maximum or minimum of function \( f(x) \) by following a three-step process:

Step One Find \( f'(x) \), set it equal to zero, and solve for x. These will be your critical points, c.

Step Two Find \( f''(x) \) and plug in the critical points. If \( f''(c) > 0 \), then \( f(c) \) is a relative minimum. If \( f''(c) < 0 \), then \( f(c) \) is a relative maximum.

Step Three Plug the critical point x-values into the original function, \(f(x)\), to find the full coordinates of the relative maxima and minima.

Step Four Determine if the extrema are absolute or relative. Absolute extrema can be determined if we are given a domain bound for the function, or if the nature of the function is such that an absolute extrema exists across all domains.

A Deeper Dive

When solving for the extrema of a function \( f(x) \), we begin by finding the first derivative and setting it equal to zero. This is due to the nature of a derivative. The derivative of a function is defined as the slope of the function. Pictured below is a function with its tangential lines graphed in blue.

Figure 3:

Figure 3: \( -x^2 + 6x - 6 \)

Thus, we see that the tangent lines at the peak of a curve have a slope of zero. This applies to minimums as well. Knowing this explains why we begin solving for the maxima and minima of a function by taking the derivative of the function and setting it equal to zero. Then, we will solve for x, which gives us the x coordinate at which a critical point occurs. This is known as the first derivative test, and it has frequent applications in optimization problems. Once we have run the first

derivative test to find the critical points for \(f(x)\), we must now determine whether these critical points are minima or maxima. We can do so by running the second derivative test. This is a test of the concavity of the function, or in which direction the function curves. We take the second derivative of \(f(x)\) and plug in the x values of the critical points. If \(f''(x) > 0\)

When given a continuous function \(f(x)\) over a bounded interval [a, b], there is both an absolute maximum and an absolute minimum. This is because the defined range between points a and b allows us to focus on a closed region of the function, allowing us to ignore the broader behavior of the function. Without a bounded interval, an absolute maximum and minimum may not exist; a visual of the graph from [−∞,∞] can help us determine whether or not the extrema are absolute.

Practice

Exercise 1. For the following function, find all critical points.

\[ f(x) = \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x \]

Step One Find \(f′(x)\).

\[ f'(x) = x^2 - 5x + 4 \]

Step Two Set \(f′(x) = 0\) and solve for x.

\[ 0 = x^2 - 5x + 4 = (x - 4)(x - 1) \]
\[ x = 1, 4 \]

Step Three Evaluate \(f(x)\) at the given x values.

\[ f(1) = \frac{1}{3}(1)^3 - \frac{5}{2}(1)^2 + 4(1) = 1.833 \]
\[ f(4) = \frac{1}{3}(4)^3 - \frac{5}{2}(4)^2 + 4(4) = -2.667 \]

Thus, our critical points are:

\[ (1, 1.833) \quad (4, -2.667) \]

We can check our critical points by graphing \(f(x)\) to see if they make sense.

Figure 4

Figure 4: \( f(x) = \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x \)

Exercise 2. Find the local minima and maxima of \( h(x) = \frac{4x}{1 + x^2} \) over \((-3,3)\). We can plot a graph of \(h(x)\) to visualize.

Figure 5

Figure 5: \( h(x) = \frac{4x}{1+x^2} \)

We can see that there is in fact a local minima and local maxima, but we will practice solving for these algebraically.

Step One Find \( h′(x) \). We will need to use the quotient rule, which is given as:

\[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x)f'(x) - f(x)g'(x)}{\left[ g(x) \right]^2} \]

Thus,

\[ h'(x) = \frac{4(1 + x^2) - 4x(2x)}{(1 + x^2)(1 + x^2)} = \frac{4 - 4x^2}{1 + 2x^2 + x^4} \]

Step Two Set \( h′(x) = 0 \) and solve for x.

\[ \frac{4 - 4x^2}{1 + 2x^2 + x^4} = 0 \]

Multiply both sides by \( 1 + 2x^2 + x^4 \) to simplify.

\[ 4 - 4x^2 = 0 \]

Thus,

\[ x = \pm 1 \]

Step Three Evaluate \(h(x)\) at these critical point x values, as well as \(h(x)\) at the endpoints \((−3, 3)\).

\[ h(-1) = \frac{4(-1)}{1 + (-1)^2} = -2 \]
\[ h(1) = \frac{4(1)}{1 + (1)^2} = 2 \]
\[ h(-3) = \frac{4(-3)}{1 + (-3)^2} = -1.2 \]
\[ h(3) = \frac{4(3)}{1 + (3)^2} = 1.2 \]

Step Four Compare the outputs. The highest value is the local maxima, and the lowest value is the local minima.

\[ h(-1) = -2 \]
\[ h(1) = 2 \]
\[ h(-3) = -1.2 \]
\[ h(3) = 1.2 \]

Thus, the local minima of \( h(x) \) is \( (−1,−2) \) and the local maxima is \( (1, 2) \).

Exercise 3. A baseball is thrown into the air and its position is given by \( g(t) = -4.9t^2 + 70t + 5 \, \text{m} \). Find the maximum height of the baseball, and the time at which this maximum height occurs. This trajectory problem can be solved in the same way as the other exercises, by finding an absolute maximum point.

Step One Find \( g′(t) \).

\[ g'(t) = -9.8t + 70 \]

Step Two Set \( g′(t) = 0 \) and solve for t.

\[ -9.8t + 70 = t \]
\[ t = 7.143 \]

Step Three Evaluate \( g(t) \) at this critical point.

\[ g(7.143) = -4.9(7.143)^2 + 70(7.143) + 5 \]
\[ g(7.143) = 255 \]

Thus, the maximum height of the baseball is 255 m, and this phenomenon occurs at \( t = 7.143 \).

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