maths.free › Calculus › 3. Using Derivatives › Applied optimization

Applied optimization

Near the conclusion of Section, we considered two optimization problems where determining the function to be optimized was part of the problem.

Introduction

Near the conclusion of Section, we considered two optimization problems where determining the function to be optimized was part of the problem. In Example, we sought to use a single piece of wire to build an equilateral triangle and square in order to maximize the total combined area enclosed. In the subsequent Activity, we investigated how the volume of a box constructed from a piece of cardboard by removing squares from each corner and folding up the sides depends on the size of the squares removed.

In neither of these problems was a function to optimize explicitly provided. Rather, we first tried to understand the problem by drawing a figure and introducing variables, and then sought to develop a formula for a function that modeled the quantity to be optimized. Once the function was established, we then considered what domain was appropriate. At that point, we were finally ready to apply the ideas of calculus to determine the absolute minimum or maximum.

Throughout what follows in the current section, the primary emphasis is on the reader solving problems. Initially, some substantial guidance is provided, with the problems progressing to require greater independence as we move along.

Exploration

Summary

While there is no single algorithm that works in every situation where optimization is used, in most of the problems we consider, the following steps are helpful: draw a picture and introduce variables; identify the quantity to be optimized and find relationships among the variables; determine a function of a single variable that models the quantity to be optimized; decide the domain on which to consider the function being optimized; use calculus to identify the absolute maximum and/or minimum of the quantity being optimized.

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Suppose you are building a fence around a rectangular field, and the field needs to have an area of 9000 square meters. There are many rectangles of different dimensions that you could consider.
1. One option you have considered is to make the field be a square. Draw a labeled picture of what the field would look like in this case.
2. Another option you have considered is to make the field look like a piece of notebook paper. Draw a picture of what the field would look like in this case, labeling the measurements of your shape. Remember, the field still needs to have an area of 9000 square meters.
3. Draw at least one more picture of a rectangular field that is different than what you've drawn before and would still have an area of 9000 square meters.
4. Which measurements are different among your three fields? Which measurements are the same?
5. Choose letters to represent the key measurements of the field. What are the lowest and highest values of those measurements?
A rectangular box with a square bottom and closed top is to be made from two materials. The material for the sides costs $1.50 per square foot and the material for the top and bottom costs $3.00 per square foot. If you are willing to spend $15 on the box, what is the largest volume it can contain? Justify your answer completely using calculus.

Jawaby görkez

Let $x$ be the length of one side of the square bottom and $h$ the height of one of the sides. We first note that the volume, $V$, of the box is $V = x^2 h$, and the surface area of the box is $S = 2x^2 + 4xh$, since there are two square sides (bottom and top) of area $x^2$ and four identical sides of area $xh$. Since the material for the top and bottom costs $3.00 per square foot and the material for the sides costs $1.50 per square foot, it follows that the total cost of a box with surface area $S$ is \[\begin{aligned}\end{aligned}\].
We are interested in building box of largest volume for which we spend $15. This cost constraint implies that \[\begin{aligned}\end{aligned}\], which allows us to express $h$ as a function of $x$ by solving for $h$. Doing so, we see that $6xh = 15 - 6x^2$, so \[\begin{aligned}\end{aligned}\]. Using this relationship, we are now able to express the volume, $V$, of the box as a function of the single variable $x$. Subsituting the expression for $h$, we have \[\begin{aligned}\end{aligned}\]. Note particularly that in order to have a box, we require that both $x \gt 0$ and $h \gt 0$. The latter condition along with the fact that $h = \frac{15-6x^2}{6x}$ shows that we need $15 - 6x^2 \gt 0$, and thus $x^2 \lt \frac{15}{6}$, so $x \lt \sqrt{\frac{15}{6}}$. Hence we are considering $V(x)$ on the domain $0 \lt x \lt \sqrt{\frac{15}{6}}$.
It is apparent that the box vanishes (and its volume tends to zero) as either $x \to 0^+$ or $x \to \sqrt{\frac{15}{6}}^-$, and thus the absolute maximum volume must appear between these values at a critical number. Differentiating $V$, we find that $V'(x) = \frac{15}{6} - 3x^2$. Solving $V'(x) = 0$ to find all critical numbers of $V$ in the domain, we see that $x^2 = \frac{15}{18}$, and thus $x = \pm \sqrt{\frac{15}{18}}$, so only $x = \sqrt{\frac{15}{18}}$ is a critical number that lies in $0 \lt x \lt \sqrt{\frac{15}{6}}$. Noting that the absolute minimum volume $V = 0$ occurs at each endpoint, it must be the case that the absolute maximum volume is $V\left( \sqrt{\frac{15}{18}} \right) = \frac{15}{6}\left( \sqrt{\frac{15}{18}} \right) - \left( \sqrt{\frac{15}{18}} \right)^3 \approx 1.52145$ cubic feet.
A farmer wants to start raising cows, horses, goats, and sheep, and desires to have a rectangular pasture for the animals to graze in. However, no two different kinds of animals can graze together. In order to minimize the amount of fencing she will need, she has decided to enclose a large rectangular area and then divide it into four equally sized pens by adding three segments of fence inside the large rectangle that are parallel to two existing sides. She has decided to purchase 7500 ft of fencing. What is the maximum possible area that each of the four pens will enclose?

Jawaby görkez

The farmer will fence a rectangular area whose length $x$ and width $y$ are unknown. The total area of the rectangular area is composed of 4 equal area rectangles. There are two ways to do this as shown in the figure below (though only the one on the left explicitly follows the way the problem is framed).
We want to maximize the area of each subrectangular region, which is the same as maximizing the total area $A = xy$ of the entire rectangular region. Since $A$ has two variables, we can't yet directly apply calculus techniques to it. We use the information about the amount of fencing to relate $x$ and $y$ in order to write $A$ in terms of just one variable.
In the case of the fencing option on the left in the figure, the total amount of fencing that is needed to enclose the four grazing areas is $2x+5y$. We assume that all of the $7500$ feet of fencing will be used, so $2x+5y=7500$. Solving for $x$, $x = 3750 - 2.5y$. Substituting this expression for $x$ into the area formula allows us to write $A$ in terms of $y$ alone: \[\begin{aligned}\end{aligned}\]. To enclose any area, we of course must have $y \gt 0$ and $x \gt 0$, and since $x = 3750 - 2.5y$, it follows that we need to have $2.5y \lt 3750$ so $y \lt 1500$. To find the maximum value of $A$ on this domain, we find the critical numbers of $A$. Since $A'(y) = 3750 - 5y$, we see that $A'(y) = 0$ when and only when $y = 750$. For $0 \lt y \lt 750$, we see that $A'(y) \gt 0$ and for $750 \lt y \lt 1500$ we have $A'(y) \lt 0$. So $A(750) = 1406250$ square feet is the absolute maximum value of $A$. The maximum possible area that each of the four pens can enclose is one quarter of the total, or $351562.5$ square feet.
In the case of the fencing option on the right in the figure, the total amount of fencing that is needed to enclose the four grazing areas is $3x+3y$. We assume that all of the $7500$ feet of fencing will be used, so $3x + 3y = 7500$. Solving for $x$ and substituting into $A$, we find $x = 2500 - y$ and \[\begin{aligned}\end{aligned}\]. To enclose any area, we of course must have $y \gt 0$ and $x \gt 0$, and since $x = 2500 - y$, it follows that we need to have $y \lt 2500$. To find the maximum value of $A$ on this domain, we find the critical numbers of $A$. Since $A'(y) = 2500 - 2y$, we see that $A'(y) = 0$ when and only when $y = 1250$. For $0 \lt y \lt 1250$, we see that $A'(y) \gt 0$ and for $1250 \lt y \lt 2500$ we have $A'(y) \lt 0$. So $A(1250) = 1406250$ square feet is the absolute maximum value of $A$. The maximum possible area that each of the four pens can enclose is one quarter of the total, or $390625$ square feet.
So the maximum possible area that each of the four pens can enclose is 390625 square feet, and we see that the righthand alternative is even more optimal than the left one.
Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched from the top of one pole to some point on the ground between the poles, and then to the top of the other pole. What is the minimum possible length of cable required? Justify your answer completely using calculus.

Jawaby görkez

We start by constructing a careful diagram of the overall situation. Let $x$ be the distance from the first pole to the second pole where the cable touches the ground, and let $y$ be the length of the cable from the top of the first tower to the ground, and $z$ the length of cable from the ground to the top of the second tower, as pictured in the following figure. Note that we have included the given heights of the towers and the distance between them.
The total amount of cable, $L$, used (which we wish to minimize) is given by $L = y + z$. We can use the geometry of the situation to write both $y$ and $z$ in terms of $x$. In the first right triangle, by the Pythagorean Theorem we know that $x^2 + 60^2 = y^2$, and thus $y = \sqrt{x^2 + 3600}$. In the second right triangle, note that the horizontal leg of the triangle has length $100-x$, and thus $(100-x)^2 + 80^2 = z^2$, so $z = \sqrt{(100-x)^2 + 6400}$. It follows that with $L = y + z$, \[\begin{aligned}\end{aligned}\]. It is also apparent from the given situation that the relevant domain for $L$ is $0 \le x \le 100$. Given that we want to minimize $L$ on this closed interval, it remains to find any critical numbers of $L$.
We note by the chain rule that \[\begin{aligned}\end{aligned}\] We use technology to solve the equation $L'(x) = 0$ and find that it has a single solution in the interval $0 \le x \le 100$: $x \approx 42.85714$.
To find the absolute minimum of $L$ on the relevant domain, we evaluate $L$ at the endpoints and the only critical number. Doing so, we fine $L(0) \approx 188.062$, $L(100) \approx 196.619$, and $L(42.85714) \approx 172.047$. Hence the least amount of cable we can use is approximately $172.047$ feet and is accomplished by anchoring the cable about $42.85714$ feet from the first tower.
A company is designing propane tanks that are cylindrical with hemispherical ends. Assume that the company wants tanks that will hold 1000 cubic feet of gas, and that the ends are more expensive to make, costing $5 per square foot, while the cylindrical barrel between the ends costs $2 per square foot. Use calculus to determine the minimum cost to construct such a tank.

Jawaby görkez

We let the length of the cylindrical part of the tank be $h$ with the radius of the cylinder given by $r$. It follows that the radius of the hemispherical ends of the tank is also $r$. We first identify formulas for both the volume and surface area of the tank by looking at the cylindrical and hemispherical pieces.
The cylindrical part of the tank has volume $V_c = \pi r^2 h$, and the surface area of its sides is $A_c = 2\pi r h$ (which is the circumference of the cylinder's base times the cylinder's height -- think about how a cylinder unfurls to form a rectangle). The two hemispherical ends together constitute a full sphere, and the volume of a sphere is $V_s = \frac{4}{3} \pi r^3$, while the surface area of a sphere is $A_s = 4 \pi r^2$.
The overall volume of the tank is therefore \[\begin{aligned}\end{aligned}\], and its total surface area is \[\begin{aligned}\end{aligned}\]. Moreover, taking into account the different costs of material for the cylindrical sides and hemispherical ends, we find that the total cost of the tank is \[\begin{aligned}\end{aligned}\].
Our goal is to minimize the total cost of the tank given that we want it to hold 1000 cubic meters of volume. From this constraint, we have the equation $1000 = \pi r^2 h + \frac{4}{3} \pi r^3$, which we can solve for $h$ to get $\pi r^2 h = 1000 - \frac{4}{3} \pi r^3$, so \[\begin{aligned}\end{aligned}\]. This allows us to now write cost as a function of the single variable $r$ by substituting the expression for $h$ into the earlier formula we found for cost. In particular, \[\begin{aligned}\end{aligned}\] Simplifying further, we have found that \[\begin{aligned}\end{aligned}\]. Based on the physical constraints of the problem, we obviously will only consider values of $r \gt 0$ and $h \gt 0$. Since $h = \frac{1000 - \frac{4}{3} \pi r^3}{\pi r^2}$, we must have \[\begin{aligned}\end{aligned}\] so $r^3 \lt \frac{3}{4 \pi} \cdot 1000 = 750$, and thus $r \lt \sqrt[3]{\frac{750}{\pi}} \approx 6.2035$. We now work to minimize $C(r)$ on the interval $0 \lt r \lt \sqrt[3]{\frac{750}{\pi}}$.
Taking the derivative of $C$, we find that $C'(r) = -\frac{4000}{r^2} + \frac{88 \pi}{3}r$. Setting $C'(r) = 0$, it follows that \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\]. Therefore, $r^3 = \frac{1500}{11 \pi}$, so $r = \sqrt[3]{\frac{1500}{11 \pi}} \approx 3.5144$ is the only critical number of $C$; observe that it lies within the interval $0 \lt r \lt \sqrt[3]{\frac{750}{\pi}} \approx 6.2035$.
If we plot the derivative function, $y = C'(r)$, on the interval $0 \lt r \lt \sqrt[3]{\frac{750}{\pi}}$, we see that for $0 \lt r \lt 3.5144$, $C'(r) \lt 0$, and for $3.5144 \lt r \lt 6.2035$, $C'(r) \gt 0$, which shows that $C$ is decreasing before the critical number and increasing after the critical number, and hence $C$ has a global minimum of $C(\sqrt[3]{\frac{1500}{11 \pi}}) \approx 1707.27$, so $1707.27 is the minimum cost to build the desired tank under the given constraints.
A rectangular box with a square bottom and closed top is to be made from two materials. The material for the sides costs $1.50 per square foot and the material for the top and bottom costs $3.00 per square foot. If you are willing to spend $15 on the box, what is the largest volume it can contain? Justify your answer completely using calculus.

Jawaby görkez

Let $x$ be the length of one side of the square bottom and $h$ the height of one of the sides. We first note that the volume, $V$, of the box is $V = x^2 h$, and the surface area of the box is $S = 2x^2 + 4xh$, since there are two square sides (bottom and top) of area $x^2$ and four identical sides of area $xh$. Since the material for the top and bottom costs $3.00 per square foot and the material for the sides costs $1.50 per square foot, it follows that the total cost of a box with surface area $S$ is \[\begin{aligned}\end{aligned}\].
We are interested in building box of largest volume for which we spend $15. This cost constraint implies that \[\begin{aligned}\end{aligned}\], which allows us to express $h$ as a function of $x$ by solving for $h$. Doing so, we see that $6xh = 15 - 6x^2$, so \[\begin{aligned}\end{aligned}\]. Using this relationship, we are now able to express the volume, $V$, of the box as a function of the single variable $x$. Subsituting the expression for $h$, we have \[\begin{aligned}\end{aligned}\]. Note particularly that in order to have a box, we require that both $x \gt 0$ and $h \gt 0$. The latter condition along with the fact that $h = \frac{15-6x^2}{6x}$ shows that we need $15 - 6x^2 \gt 0$, and thus $x^2 \lt \frac{15}{6}$, so $x \lt \sqrt{\frac{15}{6}}$. Hence we are considering $V(x)$ on the domain $0 \lt x \lt \sqrt{\frac{15}{6}}$.
It is apparent that the box vanishes (and its volume tends to zero) as either $x \to 0^+$ or $x \to \sqrt{\frac{15}{6}}^-$, and thus the absolute maximum volume must appear between these values at a critical number. Differentiating $V$, we find that $V'(x) = \frac{15}{6} - 3x^2$. Solving $V'(x) = 0$ to find all critical numbers of $V$ in the domain, we see that $x^2 = \frac{15}{18}$, and thus $x = \pm \sqrt{\frac{15}{18}}$, so only $x = \sqrt{\frac{15}{18}}$ is a critical number that lies in $0 \lt x \lt \sqrt{\frac{15}{6}}$. Noting that the absolute minimum volume $V = 0$ occurs at each endpoint, it must be the case that the absolute maximum volume is $V\left( \sqrt{\frac{15}{18}} \right) = \frac{15}{6}\left( \sqrt{\frac{15}{18}} \right) - \left( \sqrt{\frac{15}{18}} \right)^3 \approx 1.52145$ cubic feet.
A farmer wants to start raising cows, horses, goats, and sheep, and desires to have a rectangular pasture for the animals to graze in. However, no two different kinds of animals can graze together. In order to minimize the amount of fencing she will need, she has decided to enclose a large rectangular area and then divide it into four equally sized pens by adding three segments of fence inside the large rectangle that are parallel to two existing sides. She has decided to purchase 7500 ft of fencing. What is the maximum possible area that each of the four pens will enclose?

Jawaby görkez

The farmer will fence a rectangular area whose length $x$ and width $y$ are unknown. The total area of the rectangular area is composed of 4 equal area rectangles. There are two ways to do this as shown in the figure below (though only the one on the left explicitly follows the way the problem is framed).
We want to maximize the area of each subrectangular region, which is the same as maximizing the total area $A = xy$ of the entire rectangular region. Since $A$ has two variables, we can't yet directly apply calculus techniques to it. We use the information about the amount of fencing to relate $x$ and $y$ in order to write $A$ in terms of just one variable.
In the case of the fencing option on the left in the figure, the total amount of fencing that is needed to enclose the four grazing areas is $2x+5y$. We assume that all of the $7500$ feet of fencing will be used, so $2x+5y=7500$. Solving for $x$, $x = 3750 - 2.5y$. Substituting this expression for $x$ into the area formula allows us to write $A$ in terms of $y$ alone: \[\begin{aligned}\end{aligned}\]. To enclose any area, we of course must have $y \gt 0$ and $x \gt 0$, and since $x = 3750 - 2.5y$, it follows that we need to have $2.5y \lt 3750$ so $y \lt 1500$. To find the maximum value of $A$ on this domain, we find the critical numbers of $A$. Since $A'(y) = 3750 - 5y$, we see that $A'(y) = 0$ when and only when $y = 750$. For $0 \lt y \lt 750$, we see that $A'(y) \gt 0$ and for $750 \lt y \lt 1500$ we have $A'(y) \lt 0$. So $A(750) = 1406250$ square feet is the absolute maximum value of $A$. The maximum possible area that each of the four pens can enclose is one quarter of the total, or $351562.5$ square feet.
In the case of the fencing option on the right in the figure, the total amount of fencing that is needed to enclose the four grazing areas is $3x+3y$. We assume that all of the $7500$ feet of fencing will be used, so $3x + 3y = 7500$. Solving for $x$ and substituting into $A$, we find $x = 2500 - y$ and \[\begin{aligned}\end{aligned}\]. To enclose any area, we of course must have $y \gt 0$ and $x \gt 0$, and since $x = 2500 - y$, it follows that we need to have $y \lt 2500$. To find the maximum value of $A$ on this domain, we find the critical numbers of $A$. Since $A'(y) = 2500 - 2y$, we see that $A'(y) = 0$ when and only when $y = 1250$. For $0 \lt y \lt 1250$, we see that $A'(y) \gt 0$ and for $1250 \lt y \lt 2500$ we have $A'(y) \lt 0$. So $A(1250) = 1406250$ square feet is the absolute maximum value of $A$. The maximum possible area that each of the four pens can enclose is one quarter of the total, or $390625$ square feet.
So the maximum possible area that each of the four pens can enclose is 390625 square feet, and we see that the righthand alternative is even more optimal than the left one.
Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched from the top of one pole to some point on the ground between the poles, and then to the top of the other pole. What is the minimum possible length of cable required? Justify your answer completely using calculus.

Jawaby görkez

We start by constructing a careful diagram of the overall situation. Let $x$ be the distance from the first pole to the second pole where the cable touches the ground, and let $y$ be the length of the cable from the top of the first tower to the ground, and $z$ the length of cable from the ground to the top of the second tower, as pictured in the following figure. Note that we have included the given heights of the towers and the distance between them.
The total amount of cable, $L$, used (which we wish to minimize) is given by $L = y + z$. We can use the geometry of the situation to write both $y$ and $z$ in terms of $x$. In the first right triangle, by the Pythagorean Theorem we know that $x^2 + 60^2 = y^2$, and thus $y = \sqrt{x^2 + 3600}$. In the second right triangle, note that the horizontal leg of the triangle has length $100-x$, and thus $(100-x)^2 + 80^2 = z^2$, so $z = \sqrt{(100-x)^2 + 6400}$. It follows that with $L = y + z$, \[\begin{aligned}\end{aligned}\]. It is also apparent from the given situation that the relevant domain for $L$ is $0 \le x \le 100$. Given that we want to minimize $L$ on this closed interval, it remains to find any critical numbers of $L$.
We note by the chain rule that \[\begin{aligned}\end{aligned}\] We use technology to solve the equation $L'(x) = 0$ and find that it has a single solution in the interval $0 \le x \le 100$: $x \approx 42.85714$.
To find the absolute minimum of $L$ on the relevant domain, we evaluate $L$ at the endpoints and the only critical number. Doing so, we fine $L(0) \approx 188.062$, $L(100) \approx 196.619$, and $L(42.85714) \approx 172.047$. Hence the least amount of cable we can use is approximately $172.047$ feet and is accomplished by anchoring the cable about $42.85714$ feet from the first tower.
A company is designing propane tanks that are cylindrical with hemispherical ends. Assume that the company wants tanks that will hold 1000 cubic feet of gas, and that the ends are more expensive to make, costing $5 per square foot, while the cylindrical barrel between the ends costs $2 per square foot. Use calculus to determine the minimum cost to construct such a tank.

Jawaby görkez

We let the length of the cylindrical part of the tank be $h$ with the radius of the cylinder given by $r$. It follows that the radius of the hemispherical ends of the tank is also $r$. We first identify formulas for both the volume and surface area of the tank by looking at the cylindrical and hemispherical pieces.
The cylindrical part of the tank has volume $V_c = \pi r^2 h$, and the surface area of its sides is $A_c = 2\pi r h$ (which is the circumference of the cylinder's base times the cylinder's height -- think about how a cylinder unfurls to form a rectangle). The two hemispherical ends together constitute a full sphere, and the volume of a sphere is $V_s = \frac{4}{3} \pi r^3$, while the surface area of a sphere is $A_s = 4 \pi r^2$.
The overall volume of the tank is therefore \[\begin{aligned}\end{aligned}\], and its total surface area is \[\begin{aligned}\end{aligned}\]. Moreover, taking into account the different costs of material for the cylindrical sides and hemispherical ends, we find that the total cost of the tank is \[\begin{aligned}\end{aligned}\].
Our goal is to minimize the total cost of the tank given that we want it to hold 1000 cubic meters of volume. From this constraint, we have the equation $1000 = \pi r^2 h + \frac{4}{3} \pi r^3$, which we can solve for $h$ to get $\pi r^2 h = 1000 - \frac{4}{3} \pi r^3$, so \[\begin{aligned}\end{aligned}\]. This allows us to now write cost as a function of the single variable $r$ by substituting the expression for $h$ into the earlier formula we found for cost. In particular, \[\begin{aligned}\end{aligned}\] Simplifying further, we have found that \[\begin{aligned}\end{aligned}\]. Based on the physical constraints of the problem, we obviously will only consider values of $r \gt 0$ and $h \gt 0$. Since $h = \frac{1000 - \frac{4}{3} \pi r^3}{\pi r^2}$, we must have \[\begin{aligned}\end{aligned}\] so $r^3 \lt \frac{3}{4 \pi} \cdot 1000 = 750$, and thus $r \lt \sqrt[3]{\frac{750}{\pi}} \approx 6.2035$. We now work to minimize $C(r)$ on the interval $0 \lt r \lt \sqrt[3]{\frac{750}{\pi}}$.
Taking the derivative of $C$, we find that $C'(r) = -\frac{4000}{r^2} + \frac{88 \pi}{3}r$. Setting $C'(r) = 0$, it follows that \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\]. Therefore, $r^3 = \frac{1500}{11 \pi}$, so $r = \sqrt[3]{\frac{1500}{11 \pi}} \approx 3.5144$ is the only critical number of $C$; observe that it lies within the interval $0 \lt r \lt \sqrt[3]{\frac{750}{\pi}} \approx 6.2035$.
If we plot the derivative function, $y = C'(r)$, on the interval $0 \lt r \lt \sqrt[3]{\frac{750}{\pi}}$, we see that for $0 \lt r \lt 3.5144$, $C'(r) \lt 0$, and for $3.5144 \lt r \lt 6.2035$, $C'(r) \gt 0$, which shows that $C$ is decreasing before the critical number and increasing after the critical number, and hence $C$ has a global minimum of $C(\sqrt[3]{\frac{1500}{11 \pi}}) \approx 1707.27$, so $1707.27 is the minimum cost to build the desired tank under the given constraints.

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Applied optimization

In a setting where a situation is described for which optimal parameters are sought, how do we develop a function that models the situation and use calculus to find the desired maximum or minimum?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Özüňi synla

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

_Ýaşa Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests