You want to record a rocket launch, so you place your camera on your trusty tripod and get it all set up to record this event. Now, only one question remains: at what rate should your camera's angle with the ground change to allow it to keep the rocket in view as it makes its flight?
Because launching a rocket involves two related quantities that change over time, the answer to this question relies on an application of derivatives known as related rates. In this article, you will discover some of the many applications of derivatives and how they are used in calculus, engineering, and economics.
Applications of Derivatives in Calculus
Being able to solve the related rates problem discussed above is just one of many applications of derivatives you learn in calculus. You will also learn how derivatives are used to:
You will then be able to use these techniques to solve optimization problems, like maximizing an area or maximizing revenue.
Additionally, you will learn how derivatives can be applied to:
Application of Derivatives: Tangent and Normal Lines
Derivatives are very useful tools for finding the equations of tangent lines and normal lines to a curve.
Fig. 1. Using the derivative to find the tangent and normal lines to a curve.
Tangent Lines to a Curve
The tangent line to a curve is one that touches the curve at only one point and whose slope is the derivative of the curve at that point.
To find the tangent line to a curve at a given point (as in the graph above), follow these steps:
- Given a point and a curve, find the slope by taking the derivative of the given curve.
- The given point is: \[ (2, 4) \]
- The given curve is: \[ f(x) = x^{2} \]
- The derivative of the given curve is: \[ f'(x) = 2x \]
- Plug the \( x \)-coordinate of the given point into the derivative to find the slope.\[ \begin{align}f'(x) &= 2x \\f'(2) &= 2(2) \\ &= 4 \\ &= m.\end{align} \]
- Use the point-slope form of a line to write the equation.\[ \begin{align}y-y_1 &= m(x-x_1) \\y-4 &= 4(x-2) \\y &= 4(x-2)+4 \\ &= 4x - 4.\end{align} \]
For more information and examples about this subject, see our article on Tangent Lines.
Normal Lines to a Curve
The normal line to a curve is perpendicular to the tangent line. You use the tangent line to the curve to find the normal line to the curve. The slope of the normal line is:
\[ n = - \frac{1}{m} = - \frac{1}{f'(x)}. \]
To find the normal line to a curve at a given point (as in the graph above), follow these steps:
- Find the tangent line to the curve at the given point, as in the example above.
- The tangent line to the curve is: \[ y = 4(x-2)+4 \]
- Use the slope of the tangent line to find the slope of the normal line.
- The slope of the normal line to the curve is:\[ \begin{align}n &= - \frac{1}{m} \\n &= - \frac{1}{4}\end{align} \]
- Use the point-slope form of a line to write the equation.\[ \begin{align}y-y_1 &= n(x-x_1) \\y-4 &= - \frac{1}{4}(x-2) \\y &= - \frac{1}{4} (x-2)+4\end{align} \]
Application of Derivatives: Related Rates
In many real-world scenarios, related quantities change with respect to time. If you think about the rocket launch again, you can say that the rate of change of the rocket's height, \( h \), is related to the rate of change of your camera's angle with the ground, \( \theta \). In this case, you say that \( \frac{dg}{dt} \) and \( \frac{d\theta}{dt} \) are related rates because \( h \) is related to \( \theta \).
In related rates problems, you study related quantities that are changing with respect to time and learn how to calculate one rate of change if you are given another rate of change.
Strategy: Solving Related-Rates Problems
- Assign symbols to all the variables in the problem and sketch the problem if it makes sense.
- In terms of the variables you just assigned, state the information that is given and the rate of change that you need to find.
- Find an equation that relates your variables.
- Using the chain rule, take the derivative of this equation with respect to the independent variable.
- The new equation relates the derivatives.
- Substitute all the known values into the derivative, and solve for the rate of change you needed to find.
It is crucial that you do not substitute the known values too soon. If you make substitute the known values before you take the derivative, then the substituted quantities will behave as constants and their derivatives will not appear in the new equation you find in step 4.
There are lots of different articles about related rates, including Rates of Change, Motion Along a Line, Population Change, and Changes in Cost and Revenue.
Application of Derivatives: Linear Approximations and Differentials
When it comes to functions, linear functions are one of the easier ones with which to work. Therefore, they provide you a useful tool for approximating the values of other functions. You will build on this application of derivatives later as well, when you learn how to approximate functions using higher-degree polynomials while studying sequences and series, specifically when you study power series.
The key concepts and equations of linear approximations and differentials are:
A differentiable function, \( y = f(x) \), can be approximated at a point, \( a \), by the linear approximation function:
\[ L(x) = f(a) + f'(a)(x-a). \]
Given a function, \( y = f(x) \), if, instead of replacing \( x \) with \( a \), you replace \( x \) with \( a + dx \), then the differential:
\[ dy = f'(x)dx \]
is an approximation for the change in \( y \).
The actual change in \( y \), however, is:
\[ \Delta y = f(a+dx) - f(a). \]
A measurement error of \( dx \) can lead to an error in the quantity of \( f(x) \). This is known as propagated error, which is estimated by:
\[ dy \approx f'(x)dx \]
To estimate the relative error of a quantity ( \( q \) ) you use:\[ \frac{ \Delta q}{q}. \]
For more information on this topic, see our article on the Amount of Change Formula.
Application of Derivatives: Maxima and Minima
One of the most common applications of derivatives is finding the extreme values, or maxima and minima, of a function. Once you learn the methods of finding extreme values (also known collectively as extrema), you can apply these methods to later applications of derivatives, like creating accurate graphs and solving optimization problems.
The key terms and concepts of maxima and minima are:
Terms
Absolute extremum
If a function, \( f \), has an absolute max or absolute min at point \( c \), then you say that the function \( f \) has an absolute extremum at \( c \).
Absolute max / absolute maximum
If \( f(c) \geq f(x) \) for all \( x \) in the domain of \( f \), then you say that \( f \) has an absolute maximum at \( c \).
Absolute min / absolute minimum
If \( f(c) \leq f(x) \) for all \( x \) in the domain of \( f \), then you say that \( f \) has an absolute minimum at \( c \).
Local extremum
If a function, \( f \), has a local max or min at point \( c \), then you say that \( f \) has a local extremum at \( c \).
Local max / local maximum
If there exists an interval, \( I \), such that \( f(c) \geq f(x) \) for all \( x \) in \( I \), you say that \( f \) has a local max at \( c \).
Local min / local minimum
If there exists an interval, \( I \), such that \( f(c) \leq f(x) \) for all \( x \) in \( I \), you say that \( f \) has a local min at \( c \).
Critical point
Based on the definitions above, the point \( (c, f(c)) \) is a critical point of the function \( f \).
Critical number
If \( f'(c) = 0 \) or \( f'(c) \) is undefined, you say that \( c \) is a critical number of the function \( f \).
Extreme value theorem
If the function \( f \) is continuous over a finite, closed interval, then \( f \) has an absolute max and an absolute min.
Fermat's theorem
If a function \( f \) has a local extremum at point \( c \), then \( c \) is a critical point of \( f \).
Concepts
It is possible for a function to have:
both an absolute max and an absolute min,
just one absolute extremum, or
have no absolute extremum.
If a function has a local extremum, the point where it occurs must be a critical point.
A continuous function over a closed and bounded interval has an absolute max and an absolute min.
The definition of the MVT
If a function, \( f \), is continuous over the closed interval \( [a, b] \) and differentiable over the open interval \( (a, b) \), then there exists a point \( c \) in the open interval \( (a, b) \) such that
\[ f'(c) = \frac{f(b)-f(a)}{b-a}. \]
The special case of the MVT known as Rolle's theorem
If a function, \( f \), is continuous over the closed interval \( [a, b] \), differentiable over the open interval \( (a, b) \), and if \( f(a) = f(b) \), then there exists a point \( c \) in the open interval \( (a, b) \) such that
\[ f'(c) = 0 \]
The corollaries of the mean value theorem
Functions with a derivative of zero
Let \( f \) be differentiable on an interval \( I \). If \( f'(x) = 0 \) for all \( x \) in \( I \), then \( f'(x) = \) constant for all \( x \) in \( I \).
Constant difference theorem
If the functions \( f \) and \( g \) are differentiable over an interval \( I \), and \( f'(x) = g'(x) \) for all \( x \) in \( I \), then \( f(x) = g(x) + C \) for some constant \( C \).
Increasing and decreasing functions
Let \( f \) be continuous over the closed interval \( [a, b] \) and differentiable over the open interval \( (a, b) \).
If \( f'(x) > 0 \) for all \( x \) in \( (a, b) \), then \( f \) is an increasing function over \( [a, b] \).
If \( f'(x) < 0 \) for all \( x \) in \( (a, b) \), then \( f \) is a decreasing function over \( [a, b] \).
Application of Derivatives: Derivatives and the Shape of a Graph
Building on the applications of derivatives to find maxima and minima and the mean value theorem, you can now determine whether a critical point of a function corresponds to a local extreme value. But what about the shape of the function's graph? Well, this application teaches you how to use the first and second derivatives of a function to determine the shape of its graph.
Key concepts of derivatives and the shape of a graph are:
the first derivative test
Say a function, \( f \), is continuous over an interval \( I \) and contains a critical point, \( c \). If \( f \) is differentiable over \( I \), except possibly at \( c \), then \( f(c) \) satisfies one of the following:
If \( f' \) changes sign from positive when \( x < c \) to negative when \( x > c \), then \( f(c) \) is a local max of \( f \).
If \( f' \) changes sign from negative when \( x < c \) to positive when \( x > c \), then \( f(c) \) is a local min of \( f \).
If \( f' \) has the same sign for \( x < c \) and \( x > c \), then \( f(c) \) is neither a local max or a local min of \( f \).
the Candidates Test
This is a method for finding the absolute maximum and the absolute minimum of a continuous function that is defined over a closed interval. It consists of the following:
Find all the relative extrema of the function.
Evaluate the function at the extreme values of its domain.
Order the results of steps 1 and 2 from least to greatest.
test for concavity
If \( f \) is a function that is twice differentiable over an interval \( I \), then:
If \( f''(x) > 0 \) for all \( x \) in \( I \), then \( f \) is concave up over \( I \).
If \( f''(x) < 0 \) for all \( x \) in \( I \), then \( f \) is concave down over \( I \).
the second derivative test
Suppose \( f'(c) = 0 \), \( f'' \) is continuous over an interval that contains \( c \).
If \( f''(c) > 0 \), then \( f \) has a local min at \( c \).
If \( f''(c) < 0 \), then \( f \) has a local max at \( c \).
If \( f''(c) = 0 \), then the test is inconclusive.
Application of Derivatives: Limits at Infinity and Asymptotes
Once you understand derivatives and the shape of a graph, you can build on that knowledge to graph a function that is defined on an unbounded domain. To accomplish this, you need to know the behavior of the function as \( x \to \pm \infty \). This application of derivatives defines limits at infinity and explains how infinite limits affect the graph of a function.
The key terms and concepts of limits at infinity and asymptotes are:
Terms
end behavior
The behavior of the function, \( f(x) \), as \( x\to \pm \infty \).
horizontal asymptote
If \( \lim_{x \to \pm \infty} f(x) = L \), then \( y = L \) is a horizontal asymptote of the function \( f(x) \).
infinite limit at infinity
The function \( f(x) \) becomes larger and larger as \( x \) also becomes larger and larger.
limit at infinity
The limiting value, if it exists, of a function \( f(x) \) as \( x \to \pm \infty \).
oblique asymptote
The line \( y = mx + b \), if \( f(x) \) approaches it, as \( x \to \pm \infty \) is an oblique asymptote of the function \( f(x) \).
Concepts
The limit of the function \( f(x) \) is \( L \) as \( x \to \pm \infty \) if the values of \( f(x) \) get closer and closer to \( L \) as \( x \) becomes larger and larger.
The limit of the function \( f(x) \) is \( \infty \) as \( x \to \infty \) if \( f(x) \) becomes larger and larger as \( x \) also becomes larger and larger.
The limit of the function \( f(x) \) is \( - \infty \) as \( x \to \infty \) if \( f(x) < 0 \) and \( \left| f(x) \right| \) becomes larger and larger as \( x \) also becomes larger and larger.
For the polynomial function \( P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{1}x + a_{0} \), where \( a_{n} \neq 0 \), the end behavior is determined by the leading term: \( a_{n}x^{n} \).
For the rational function \( f(x) = \frac{p(x)}{q(x)} \), the end behavior is determined by the relationship between the degree of \( p(x) \) and the degree of \( q(x) \).
If the degree of \( p(x) \) is less than the degree of \( q(x) \), then the line \( y = 0 \) is a horizontal asymptote for the rational function.
If the degree of \( p(x) \) is equal to the degree of \( q(x) \), then the line \( y = \frac{a_{n}}{b_{n}} \), where \( a_{n} \) is the leading coefficient of \( p(x) \) and \( b_{n} \) is the leading coefficient of \( q(x) \), is a horizontal asymptote for the rational function.
If the degree of \( p(x) \) is greater than the degree of \( q(x) \), then the function \( f(x) \) approaches either \( \infty \) or \( - \infty \) at each end.
Application of Derivatives: Optimization Problems
Continuing to build on the applications of derivatives you have learned so far, optimization problems are one of the most common applications in calculus. These are defined as calculus problems where you want to solve for a maximum or minimum value of a function.
Strategy: Solving Optimization Problems
- Introduce all variables.
- If it makes sense, draw a figure and label all your variables.
- Determine which quantity (which of your variables from step 1) you need to maximize or minimize.
- Determine for what range of values of the other variables (if this can be determined at this time) you need to maximize or minimize your quantity.
- Write a formula for the quantity you need to maximize or minimize in terms of your variables.
- This formula will most likely involve more than one variable.
- Write any equations you need to relate the independent variables in the formula from step 3.
- Use these equations to write the quantity to be maximized or minimized as a function of one variable.
- Identify the domain of consideration for the function in step 4.
- Make sure you consider the physical problem to be solved.
- Locate the maximum or minimum value of the function from step 4.
- This step usually involves looking for critical points and evaluating a function at endpoints.
Application of Derivatives: L’Hôpital’s Rule
A powerful tool for evaluating limits, L’Hôpital’s Rule is yet another application of derivatives in calculus. This application uses derivatives to calculate limits that would otherwise be impossible to find. These limits are in what is called indeterminate forms.
The key terms and concepts of L’Hôpital’s Rule are:
Terms
indeterminate forms
When evaluating a limit, the forms \[ \frac{0}{0}, \ \frac{\infty}{\infty}, \ 0 \cdot \infty, \ \infty - \infty, \ 0^{0}, \ \infty^{0}, \ \mbox{ and } 1^{\infty} \] are all considered indeterminate forms because you need to further analyze (i.e., by using L’Hôpital’s rule) whether the limit exists and, if so, what the value of the limit is.
L’Hôpital’s Rule
If two functions, \( f(x) \) and \( g(x) \), are differentiable functions over an interval \( a \), except possibly at \( a \), and \[ \lim_{x \to a} f(x) = 0 = \lim_{x \to a} g(x) \] or \[ \lim_{x \to a} f(x) \mbox{ and } \lim_{x \to a} g(x) \mbox{ are infinite, } \] then \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \] assuming the limit involving \( f'(x) \) and \( g'(x) \) either exists or is \( \pm \infty \).
Concepts
You can use L’Hôpital’s rule to evaluate the limit of a quotient when it is in either of the indeterminate forms \( \frac{0}{0}, \ \frac{\infty}{\infty} \).
You can also use L’Hôpital’s rule on the other indeterminate forms if you can rewrite them in terms of a limit involving a quotient when it is in either of the indeterminate forms \( \frac{0}{0}, \ \frac{\infty}{\infty} \).
Application of Derivatives: Newton’s Method
In many applications of math, you need to find the zeros of functions. Unfortunately, it is usually very difficult – if not impossible – to explicitly calculate the zeros of these functions. Newton's method saves the day in these situations because it is a technique that is efficient at approximating the zeros of functions.
The key terms and concepts of Newton's method are:
Terms
iterative process
A process in which a list of numbers like \[ x_{0}, x_{1}, x_{2}, \ldots \] is generated by beginning with a number \( x_{0} \) and then defining \[ x_{n} = F \left( x_{n-1} \right) \] for \( n \neq 1 \).
Newton's method
A method for approximating the roots of \( f(x) = 0 \). It uses an initial guess of \( x_{0} \). Each subsequent approximation is defined by the equation \[ x_{n} = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}. \]
Concepts
Newton's method approximates the roots of \( f(x) = 0 \) by starting with an initial approximation of \( x_{0} \). From there, it uses tangent lines to the graph of \( f(x) \) to create a sequence of approximations \( x_1, x_2, x_3, \ldots \).
Failures of Newton's method:
This method fails when the list of numbers \( x_1, x_2, x_3, \ldots \) does not approach a finite value, or
when it approaches a value other than the root you are looking for.
Any process in which a list of numbers \( x_1, x_2, x_3, \ldots \) is generated by defining an initial number \( x_{0} \) and defining the subsequent numbers by the equation \[ x_{n} = F \left( x_{n-1} \right) \] for \( n \neq 1 \) is an iterative process.
Application of Derivatives: Antiderivatives
Having gone through all the applications of derivatives above, now you might be wondering: what about turning the derivative process around? What if I have a function \( f(x) \) and I need to find a function whose derivative is \( f(x) \)? What application does this have?
To answer these questions, you must first define antiderivatives.
One of many examples where you would be interested in an antiderivative of a function is the study of motion.
The key terms and concepts of antiderivatives are:
Terms
antiderivative
A function \( F(x) \) such that \( F'(x) = f(x) \) for all \( x \) in the domain of \( f \) is an antiderivative of \( f \).
indefinite integral
The most general antiderivative of a function \( f(x) \) is the indefinite integral of \( f \). The notation \[ \int f(x) dx \] denotes the indefinite integral of \( f(x) \).
initial value problem
A problem that requires you to find a function \( y \) that satisfies the differential equation \[ \frac{dy}{dx} = f(x) \] together with the initial condition of \[ y(x_{0}) = y_{0}. \]
Concepts
If the function \( F \) is an antiderivative of another function \( f \), then every antiderivative of \( f \) is of the form \[ F(x) + C \] for some constant \( C \).
Solving the initial value problem \[ \frac{dy}{dx} = f(x), \mbox{ with the initial condition } y(x_{0}) = y_{0} \] requires you to:
Applications of Derivatives in Engineering
The applications of derivatives in engineering is really quite vast. To touch on the subject, you must first understand that there are many kinds of engineering. To name a few;
Mechanical Engineering
Civil Engineering
Industrial Engineering
Electrical Engineering
Aerospace Engineering
Chemical Engineering
Computer Engineering
\( \vdots \)
All of these engineering fields use calculus. They all use applications of derivatives in their own way, to solve their problems.
In every case, to study the forces that act on different objects, or in different situations, the engineer needs to use applications of derivatives (and much more).
Applications of Derivatives in Economics
Even the financial sector needs to use calculus! Applications of derivatives are used in economics to determine and optimize:
supply and demand,
profit and cost, and
revenue and loss.
Applications of Derivatives: Examples
Launching a Rocket – Related Rates Example
Your camera is set up \( 4000ft \) from a rocket launch pad. The rocket launches, and when it reaches an altitude of \( 1500ft \) its velocity is \( 500ft/s \). What rate should your camera's angle with the ground change to allow it to keep the rocket in view as it makes its flight?
Solution:
- Sketch the problem.
Your camera is \( 4000ft \) from the launch pad of a rocket. The two related rates – the angle of your camera \( (\theta) \) and the height \( (h) \) of the rocket – are changing with respect to time \( (t) \).
- Here, \( \theta \) is the angle between your camera lens and the ground and \( h \) is the height of the rocket above the ground.
- Clarify what exactly you are trying to find.
- The problem asks you to find the rate of change of your camera's angle to the ground when the rocket is \( 1500ft \) above the ground. Both of these variables are changing with respect to time.
- This means you need to find \( \frac{d \theta}{dt} \) when \( h = 1500ft \). You also know that the velocity of the rocket at that time is \( \frac{dh}{dt} = 500ft/s \).
- Determine what equation relates the two quantities \( h \) and \( \theta \).
- Looking back at your picture in step \( 1 \), you might think about using a trigonometric equation. What relates the opposite and adjacent sides of a right triangle? The \( \tan \) function! So, you have:\[ \tan(\theta) = \frac{h}{4000} .\]
- Rearranging to solve for \( h \) gives:\[ h = 4000\tan(\theta). \]
- Differentiate this to get:\[ \frac{dh}{dt} = 4000\sec^{2}(\theta)\frac{d\theta}{dt} .\]
- Find \( \frac{d \theta}{dt} \) when \( h = 1500ft \).
- To find \( \frac{d \theta}{dt} \), you first need to find \(\sec^{2} (\theta) \). How can you do that?
- Going back to trig, you know that \( \sec(\theta) = \frac{\text{hypotenuse}}{\text{adjacent}} \).
- And, from the givens in this problem, you know that \( \text{adjacent} = 4000ft \) and \( \text{opposite} = h = 1500ft \).
- So, you can use the Pythagorean theorem to solve for \( \text{hypotenuse} \).\[ \begin{align}a^{2}+b^{2} &= c^{2} \\(4000)^{2}+(1500)^{2} &= (\text{hypotenuse})^{2} \\\text{hypotenuse} &= 500 \sqrt{73}ft.\end{align} \]
- Therefore, when \( h = 1500ft \), \( \sec^{2} ( \theta ) \) is:\[ \begin{align}\sec^{2}(\theta) &= \left( \frac{\text{hypotenuse}}{\text{adjacent}} \right)^{2} \\&= \left( \frac{500 \sqrt{73}}{4000} \right)^{2} \\&= \frac{73}{64}.\end{align} \]
- Plug in the values for \( \sec^{2}(\theta) \) and \( \frac{dh}{dt} \) into the function you found in step 4 and solve for \( \frac{d \theta}{dt} \).\[ \begin{align}\frac{dh}{dt} &= 4000\sec^{2}(\theta)\frac{d\theta}{dt} \\500 &= 4000 \left( \frac{73}{64} \right) \frac{d\theta}{dt} \\\frac{d\theta}{dt} &= \frac{8}{73}.\end{align} \]
- Therefore, the rate that your camera's angle with the ground should change to allow it to keep the rocket in view as it makes its flight is:\[ \frac{d\theta}{dt} = \frac{8}{73} rad/s. \]
Engineering Application – Optimization Example
You are an agricultural engineer, and you need to fence a rectangular area of some farmland. One side of the space is blocked by a rock wall, so you only need fencing for three sides. Given that you only have \( 1000ft \) of fencing, what are the dimensions that would allow you to fence the maximum area? What is the maximum area?
Determine the dimensions \( x \) and \( y \) that will maximize the area of the farmland using \( 1000ft \) of fencing.
Solution:
- Let \( x \) be the length of the sides of the farmland that run perpendicular to the rock wall, and let \( y \) be the length of the side of the farmland that runs parallel to the rock wall. Then the area of the farmland is given by the equation for the area of a rectangle:\[ A = x \cdot y. \]
- Since you want to find the maximum possible area given the constraint of \( 1000ft \) of fencing to go around the perimeter of the farmland, you need an equation for the perimeter of the rectangular space.
- Don't forget to consider that the fence only needs to go around \( 3 \) of the \( 4 \) sides! So, your constraint equation is:\[ 2x + y = 1000. \]
- Now, you want to solve this equation for \( y \) so that you can rewrite the area equation in terms of \( x \) only:\[ y = 1000 - 2x. \]
- Rewriting the area equation, you get:\[ \begin{align}A &= x \cdot y \\A &= x \cdot (1000 - 2x) \\A &= 1000x - 2x^{2}.\end{align} \]
- Before jumping right into maximizing the area, you need to determine what your domain is.
- First, you know that the lengths of the sides of your farmland must be positive, i.e., \( x \) and \( y \) can't be negative numbers.
- Since \( y = 1000 - 2x \), and you need \( x > 0 \) and \( y > 0 \), then when you solve for \( x \), you get:\[ x = \frac{1000 - y}{2}. \]
- Minimizing \( y \), i.e., if \( y = 1 \), you know that:\[ x < 500. \]
- So, you need to determine the maximum value of \( A(x) \) for \( x \) on the open interval of \( (0, 500) \).
- However, you don't know that a function necessarily has a maximum value on an open interval, but you do know that a function does have a max (and min) value on a closed interval. Therefore, you need to consider the area function \( A(x) = 1000x - 2x^{2} \) over the closed interval of \( [0, 500] \).
- Find the max possible area of the farmland by maximizing \( A(x) = 1000x - 2x^{2} \) over the closed interval of \( [0, 500] \).
- Since \( A(x) \) is a continuous function on a closed, bounded interval, you know that, by the extreme value theorem, it will have maximum and minimum values. These extreme values occur at the endpoints and any critical points.
- At the endpoints, you know that \( A(x) = 0 \).
- Since the area must be positive for all values of \( x \) in the open interval of \( (0, 500) \), the max must occur at a critical point. To find critical points, you need to take the first derivative of \( A(x) \), set it equal to zero, and solve for \( x \).\[ \begin{align}A(x) &= 1000x - 2x^{2} \\A'(x) &= 1000 - 4x \\0 &= 1000 - 4x \\x &= 250.\end{align} \]
- The only critical point is \( x = 250 \). Therefore, the maximum area must be when \( x = 250 \).
- Plugging this value into your perimeter equation, you get the \( y \)-value of this critical point:\[ \begin{align}y &= 1000 - 2x \\y &= 1000 - 2(250) \\y &= 500.\end{align} \]
- Therefore, to maximize the area of the farmland, \( x = 250ft \) and \( y = 500ft \). The area is \( 125000ft^{2} \). The graph below visualizes this.
To maximize the area of the farmland, you need to find the maximum value of \( A(x) = 1000x - 2x^{2} \).
Economic Application – Optimization Example
You are the Chief Financial Officer of a rental car company. You found that if you charge your customers \( p \) dollars per day to rent a car, where \( 20 < p < 100 \), the number of cars \( n \) that your company rent per day can be modeled using the linear function
\[ n(p) = 600 - 6p. \]
If the company charges \( $20 \) or less per day, they will rent all of their cars. If the company charges \( $100 \) per day or more, they won't rent any cars.
How much should you tell the owners of the company to rent the cars to maximize revenue?
Solution:
- Let \( p \) be the price charged per rental car per day. Let \( n \) be the number of cars your company rents per day. Let \( R \) be the revenue earned per day.
- Find an equation that relates all three of these variables.
- Revenue earned per day is the number of cars rented per day times the price charged per rental car per day:\[ R = n \cdot p. \]
- Substitute the value for \( n \) as given in the original problem.\[ \begin{align}R &= n \cdot p \\R &= (600 - 6p)p \\R &= -6p^{2} + 600p.\end{align} \]
- Determine what your domain is.
- Since you intend to tell the owners to charge between \( $20 \) and \( $100 \) per car per day, you need to find the maximum revenue for \( p \) on the closed interval of \( [20, 100] \).
- Find the maximum possible revenue by maximizing \( R(p) = -6p^{2} + 600p \) over the closed interval of \( [20, 100] \).
- Since \( R(p) \) is a continuous function over a closed, bounded interval, you know that, by the extreme value theorem, it will have maximum and minimum values. These extreme values occur at the endpoints and any critical points.
- Find the critical points by taking the first derivative, setting it equal to zero, and solving for \( p \).\[ \begin{align}R(p) &= -6p^{2} + 600p \\R'(p) &= -12p + 600 \\0 &= -12p + 600 \\p = 50.\end{align} \]
- The only critical point is \( p = 50 \). Therefore, the maximum revenue must be when \( p = 50 \).
- Plugging this value into your revenue equation, you get the \( R(p) \)-value of this critical point:\[ \begin{align}R(p) &= -6p^{2} + 600p \\R(50) &= -6(50)^{2} + 600(50) \\R(50) &= 15000.\end{align} \]
- Therefore, to maximize revenue, you should tell the owners to charge \( $50 \) per car per day. The graph below visualizes this.
To maximize revenue, you need to balance the price charged per rental car per day against the number of cars customers will rent at that price.
Application of Derivatives – Key takeaways
- In calculus, there are many applications of derivatives, including:
- Derivatives are useful beyond the realm of math, in fields like:
- Engineering
- Physics
- Economics
- Business
- Health
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