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Maxima and Minima Problems

- Calculus
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Imagine, if you will, sitting in a classroom working out calculus problems on your test. And then, you come to it, the bane of many a calculus student: the dreaded maxima and minima word problem.

There is usually only one of these problems on your test – so you have only one shot to prove your skills – because they are so lengthy. Why are they like this? Of what use are they, really?

Well, just a few common applications of maxima and minima problems are:

to minimize cost and maximize profit,

to minimize or maximize areas,

to determine maximum or minimum values of an object in motion,

to help determine the dosage of a medicine, and

to help determine which materials to use when engineering something.

Now, if you found your way here from the maxima and minima article, great! If not, please refer to the maxima and minima article for a more in-depth introduction to the topic.

That being said, in this article, you will work through many examples of the application of derivatives known as maxima and minima problems.

First and foremost, what do maxima and minima look like?

**Maxima** are where a function has a high point, sometimes called a peak.

In turn,

**Minima** are where a function has a low point, sometimes called a valley.

Both maxima and minima can be included in a single term that is **extrema**.

Look at figure 1 to better clarify these terms.

Fig. 1 - The graph of the function \( f(x) = 2x^{3}-x^{5} \) has one local minimum, a saddle point, and one local maximum.

In a smooth function (i.e., one with no sharp points, breaks, or discontinuities) a maximum or minimum value is always where the function flattens out – except for a saddle point.

Before moving on, there are a couple of points of note regarding this graph.

**First:**

A **saddle point** is a critical point of a function that is neither a maximum nor a minimum.

In other words, a saddle point is a point on a function where *the derivative is zero*, but the point is **not** the highest or lowest in its area.

And, often, a saddle point looks much like the saddle on a horse. Hence, the name.

**Second:**

If you look carefully, you will see that the graph above does not have any absolute maxima or minima. This is because the function has a domain of \( (-\infty, \infty) \) and it extends towards infinity on both sides. To express this more formally,

*a function that is defined on an open interval and whose left- and right-hand sides extend toward either positive or negative infinity has no absolute extrema*.

For a more in-depth analysis and explanation of absolute extrema, please refer to the article on absolute maxima and minima.

Now that you know what maxima and minima look like, let's review the steps in solving maxima and minima problems. To do this, we will organize the steps into Strategy 1 and Strategy 2.

Let's start by looking at the first strategy.

**Strategy 1 – Finding the Relative/Local Extrema of a Function**

- Take the first derivative of the given function.
- Set \( f'(x) = 0 \) and solve for \( x \) to find all critical points.
- Take the second derivative of the given function.
- Plug in the critical points from step \( 2 \) into the second derivative.
- If \( f''(c) < 0 \), then the critical point of \( f(x) \) is a maximum.
- If \( f''(c) > 0 \), then the critical point of \( f(x) \) is a minimum.

Now, let's move on to the second strategy.

**Strategy 2 – Finding the Absolute/Global Extrema of a Function**

To find the absolute extrema of a function, it must be continuous and defined over a closed interval \( [a, b] \).

- Solve the function at its endpoints, i.e., where \( x = a \) and \( x = b \).
- Find all the critical points of the function that are on the open interval \( (a, b) \) and solve the function at each critical point.
- Take the first derivative of the given function.
- Set \( f'(x) = 0 \) and solve for \( x \) to find all critical points.
- Take the second derivative of the given function.
- Plug in the critical points from step 2 into the second derivative.
- If \( f''(c) < 0 \), then the critical point of \( f(x) \) is a maximum.
- If \( f''(c) > 0 \), then the critical point of \( f(x) \) is a minimum.

- Compare all the values from steps \( 1 \) and \( 2 \).
- The largest of the values is the absolute maximum of the function.
- The smallest of the values is the absolute minimum of the function.
- All other values are relative/local extrema of the function.

In this section, you will work through examples where you identify the extrema of a given function graph.

Identify all the maxima and minima of the graph. You should assume that the graph represents the entire function, that is, it is on a closed interval.

**Solution**:

We can solve this example just by looking at the graph.

\( x \) | \( f(x) \) | Conclusion |

\( -3 \) | \( -1 \) | relative min |

\( -1.5 \) | \( 12 \) | absolute max |

\( 1.5 \) | \( -2 \) | absolute min |

\( 3 \) | \( 11 \) | relative max |

Let's look at another example.

Identify all the maxima and minima of the graph. You should assume that the graph represents the entire function, that is, it is on a closed interval.

**Solution**:

\( x \) | \( f(x) \) | Conclusion |

\( -3 \) | \( 66 \) | absolute max |

\( 0.25 \) | \( 3 \) | absolute min |

\( 1.4 \) | \( 8 \) | relative max |

\( 2.1 \) | \( 4 \) | relative min |

\( 2.75 \) | \( 13 \) | relative max |

\( 3 \) | \( 6 \) | relative min |

In this section, you will work through examples where you find the extrema of a given function.

Find the local extrema (maxima and minima) for the function,

\[ f(x) = x^{3} - 3x + 5. \]

**Solution**:

Here, you must apply Strategy 1.

Take the first derivative of the given function.\[ \begin{align}f'(x) &= \frac{\mathrm{d}}{\mathrm{d}x}(x^{3} - 3x + 5) \\ \\f'(x) &= 3x^{2} - 3\end{align} \]

Set \( f'(x) = 0 \) and solve for \( x \) to find all critical points.\[ \begin{align}f'(x) = 3x^{2} - 3 &= 0 \\ \\3(x^{2} - 1) &= 0 \\ \\(x - 1)(x + 1) &= 0 \\ \\x &= \pm 1\end{align} \]

Take the second derivative of the given function.\[ \begin{align}f''(x) = \frac{\mathrm{d}}{\mathrm{d}x}(f'(x)) &= \frac{\mathrm{d}}{\mathrm{d}x}(3x^{2} - 3) \\ \\f''(x) &= 6x\end{align} \]

Plug in the critical points from step 2 into the second derivative.

For \( x = -1 \),\[ f''(x) = 6(-1) = -6 \]Since \( -6 \) is negative, the critical point where \( x = -1 \) is a

**local maximum**.For \( x = 1 \),\[ f''(x) = 6(1) = 6 \]Since \( 6 \) is positive, the critical point where \( x = 1 \) is a

**local minimum**.

To confirm your calculations, graph the function and plot the extreme values.

**Therefore, the point \( (-1, 7) \) is a local maximum and the point \( (1, 3) \) is a local minimum.**

Let's explore another example.

Find the absolute extrema (maximum and minimum) of the function,

\[ f(x) = x^{3} - 3x + 5 \]

over the closed interval \( [-3, 3] \).

**Solution**:

Here you must apply Strategy 2.

- Solve the function at its endpoints, i.e., where \( x = -3 \) and \( x = 3 \).\[ \begin{align}f(-3) &= (-3)^{3} - 3(-3) + 5 = -13 \\ \\f(3) &= (3)^{3} - 3(3) + 5 = 23\end{align} \]
- Find all the critical points of the function that are on the open interval \( (-3, 3) \) and solve the function at each critical point.
Since this is the same function that you used in the previous example, please refer to steps \( 1 \) through \( 4 \) of that example.

- Compare all the values from steps \( 1 \) and \( 2 \).

\( x \) | \( f(x) \) |

\( -3 \) | \( -13 \) |

\( -1 \) | \( 7 \) |

\( 1 \) | \( 3 \) |

\( 3 \) | \( 23 \) |

To confirm your calculations and comparisons, graph the function and plot the extrema.

**Therefore, the point \( (-3, -13) \) is the absolute minimum and the point \( (3, 23) \) is the absolute maximum.**

In this section – like the previous one – you will work through examples where you find the extrema of a function as a real-world application of taking derivatives. The difference here is how the information is presented to you. Following the format of a word problem, you will have to be able to determine what you need to do based on the context.

Say you launch a model rocket, and you know that the height of the rocket with respect to time is given by the formula

\[ H(t) = -6t^{2} + 120t \]

where,

- \( H \) is in meters,
- \( t \) is in seconds, and
- \( t > 0 \).

- How long does the model rocket take to reach its maximum height?
- What is the maximum height the model rocket reaches?
- How long does the model rocket take to land back on the ground?

**Solutions**:

Based on the given information, you know that:

- the height of the model rocket is given by the variable \( H \),
- the time elapsed – in seconds – is given by the variable \( t \), and
- the time can't be negative.

- This is asking you to find the maximum value of the function \( H(t) \).
- Take the first derivative of the given function.\[ \begin{align}H'(t) &= \frac{\mathrm{d}}{\mathrm{d}x}(-6t^{2} + 120t) \\ \\H'(t) &= -12t + 120\end{align} \]
- Set \( H'(t) = 0 \) and solve for \( t \) to find all critical points.\[ \begin{align}H'(t) = -12t + 120 &= 0 \\ \\-12t &= -120 \\ \\t &= 10\end{align} \]
- Take the second derivative of the given function.\[ \begin{align}H''(t) &= \frac{\mathrm{d}}{\mathrm{d}x}(-12t + 120) \\ \\H''(t) &= -12\end{align} \]
- Plug in the critical points from step \( 2 \) into the second derivative.\[ H''(10) = -12 \]
- Since \( H''(10) \) is negative, \( t = 10 \) is the maximum value of the function.
**Therefore, \( t = 10\space s \) is how long the model rocket takes to reach its maximum height**.

- Since \( H''(10) \) is negative, \( t = 10 \) is the maximum value of the function.

- To find the maximum height the model rocket reaches, you take the time at which the model rocket reaches its maximum height, which you found in part A, and plug it into the given function \( H(t) \).\[ \begin{align}H(t) &= -6t^{2} + 120t & \mbox{ the given function } \\ \\H(10) &= -6(10)^{2} + 120(10) & \mbox { plug in 10 for t } \\ \\&= -6(100) +1200 & \mbox{ simplify } \\ \\&= -600 + 1200 \\ \\&= 600\end{align} \]
**Therefore, \( H = 600\space m \) is the maximum height the model rocket reaches**. - To find when the model rocket lands on the ground after being launched, you need to think about what is physically happening.
- There are two cases when the model rocket is on the ground in this scenario:
- when the model rocket is launched, and
- when it lands.

- What is common in both of these cases?
- The height of the model rocket is \( 0 \)!
- So, to answer this question, you need to set the original function \( H(t) \) equal to \( 0 \) and solve for \( t \).\[ \begin{align}H(t) = -6t^{2} + 120t &= 0 \\ \\-6t(t - 20) &= 0 \\ \\t = 0 \mbox{ and } t = 20\end{align} \]
- You know that when \( t = 0\space s \) is when the model rocket was launched initially, so that means that
**when \( t = 20\space s \) is how long the model rocket takes to land after being launched**.

- There are two cases when the model rocket is on the ground in this scenario:

Let's consider another example.

A manufacturing company finds that its profit from assembling a certain number of bicycles per day is given by the formula

\[ P(n) = -n^{2} + 50n - 100. \]

- How many bicycles need to be assembled per day to maximize profit?
- What is the maximum profit?
- What is the loss if no bicycles are assembled in one day?

**Solutions**:

Based on the given information, you know that:

- the profit the manufacturing company makes – in dollars – is given by the variable \( P \),
- the number of bicycles assembled in one day is given by the variable \( n \), and
- the number of bicycles assembled can't be negative.

- This is asking you to find the maximum value of the function \( P(n) \).
- Take the first derivative of the given function.\[ \begin{align}P'(n) &= \frac{\mathrm{d}}{\mathrm{d}x}(-n^{2} + 50n - 100) \\ \\P'(n) &= -2n + 50\end{align} \]
- Set \( P'(n) = 0 \) and solve for \( n \) to find all critical points.\[ \begin{align}P'(n) = -2n + 50 &= 0 \\ \\-2n &= -50 \\ \\n &= 25\end{align} \]
- Take the second derivative of the given function.\[ \begin{align}P''(n) &= \frac{\mathrm{d}}{\mathrm{d}x}(-2n + 50) \\ \\P''(n) &= -2\end{align} \]
- Plug in the critical points from step \( 2 \) into the second derivative.\[ P''(25) = -2 \]
- Since \( P''(25) \) is negative, \( n = 25 \) is the maximum value of the function.
**Therefore, \( n = 25 \) is the number of bicycles that need to be assembled per day to maximize profit**.

- Since \( P''(25) \) is negative, \( n = 25 \) is the maximum value of the function.

- To find the maximum profit, you take the number of bicycles that must be assembled per day to maximize profit, which you found in part A, and plug it into the given function \( P(n) \).\[ \begin{align}P(n) &= -n^{2} + 50n - 100 & \mbox{ the given function } \\ \\P(25) &= -(25)^{2} + 50(25) - 100 & \mbox { plug in 25 for n } \\ \\&= -(625) + 1250 - 100 & \mbox{ simplify } \\ \\&= -625 + 1150 \\ \\&= 525\end{align} \]
**Therefore, \( P = $525 \) is the maximum profit**. - What do you know if no bicycles are assembled in one day? That's right, \( n = 0 \)! So, to find the loss if no bicycles are assembled in one day, you solve the given function when \( n = 0 \).\[ \begin{align}P(n) &= -n^{2} + 50n - 100 & \mbox{ the given function } \\ \\P(0) &= -(0)^{2} + 50(0) - 100 & \mbox { plug in 0 for n } \\ \\&= -(0) + 0 - 100 & \mbox{ simplify } \\ \\&= -100\end{align} \]
**Therefore, \( P = -$100 \) is the loss if no bicycles are assembled in one day**.

- Extreme values of a function are collectively known as
**extrema**. Extrema are also called**maxima and minima**. - Maxima and minima are the peaks and valleys on the graph of a function.
- A function can have only one absolute maximum and one absolute minimum on its domain.
- Common applications of maxima and minima problems are:
- to minimize cost and maximize profit
- to minimize or maximize areas
- to determine maximum or minimum values of an object in motion
- to help determine the dosage of a medicine
- to help determine which materials to use when engineering something

You solve maxima and minima problems by using the first and second derivatives of the original function. These are used in what are called the first derivative test and the second derivative test.

- The first derivative test tells you where a max or min point exists along the original function by telling you where the slope of the function is zero.
- The second derivative test tells you if the point found in the first derivative test is a max or a min.

Maxima is the plural of maximum.

- The maximum of a function is its largest output.
- A parabola that opens downward has its maximum at its vertex.

Minima is the plural of minimum.

- The minimum of a function is its smallest output.
- A parabola that opens upward has its minimum at its vertex.

More about Maxima and Minima Problems

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