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Applied Optimization Problems

Set up and solve optimization problems in several applied fields.

Solving Optimization Problems over a Closed, Bounded Interval

The basic idea of the optimization problems that follow is the same. We have a particular quantity that we are interested in maximizing or minimizing. However, we also have some auxiliary condition that needs to be satisfied. For example, in , we are interested in maximizing the area of a rectangular garden. Certainly, if we keep making the side lengths of the garden larger, the area will continue to become larger. However, what if we have some restriction on how much fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Let’s look at how we can maximize the area of a rectangle subject to some constraint on the perimeter.

Example

Try it.

A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides (). Given $100$ ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?

Solution

Let $x$ denote the length of the side of the garden perpendicular to the rock wall and $y$ denote the length of the side parallel to the rock wall. Then the area of the garden is

\[A=x\cdot y.\]

We want to find the maximum possible area subject to the constraint that the total fencing is $100\ \text{ft}.$ From , the total amount of fencing used will be $2x+y.$ Therefore, the constraint equation is

\[2x+y=100.\]

Solving this equation for $y,$ we have $y=100-2x.$ Thus, we can write the area as

\[A(x)=x\cdot (100-2x)=100x-2{x}^{2}.\]

Before trying to maximize the area function $A(x)=100x-2{x}^{2},$ we need to determine the domain under consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need $x>0$ and $y>0.$ Since $y=100-2x,$ if $y>0,$ then $x<50.$ Therefore, we are trying to determine the maximum value of $A(x)$ for $x$ over the open interval $(0,50).$ We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval. Therefore, let’s consider the function $A(x)=100x-2{x}^{2}$ over the closed interval $[0,50].$ If the maximum value occurs at an interior point, then we have found the value $x$ in the open interval $(0,50)$ that maximizes the area of the garden. Therefore, we consider the following problem:

Maximize $A(x)=100x-2{x}^{2}$ over the interval $[0,50].$

As mentioned earlier, since $A$ is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, $A(x)=0.$ Since the area is positive for all $x$ in the open interval $(0,50),$ the maximum must occur at a critical point. Differentiating the function $A(x),$ we obtain

\[{A}^{'}(x)=100-4x.\]

Therefore, the only critical point is $x=25$ (). We conclude that the maximum area must occur when $x=25.$ Then we have $y=100-2x=100-2(25)=50.$ To maximize the area of the garden, let $x=25$ ft and $y=50\ \text{ft}.$ The area of this garden is $1250{\ \text{ft}}^{2}.$

Now let’s look at a general strategy for solving optimization problems similar to .

Condensed — the full section is in OpenStax Calculus Volume 1.

Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded

In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain is neither closed nor bounded.

Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function $f(x)={x}^{2}+4$ over $(\text{-}\infty ,\infty )$ has an absolute minimum of $4$ at $x=0.$ Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of consideration is $(0,\infty ),$ the function has an absolute minimum.

In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

To solve an optimization problem, begin by drawing a picture and introducing variables.
Find an equation relating the variables.
Find a function of one variable to describe the quantity that is to be minimized or maximized.
Look for critical points to locate local extrema.

Applied Optimization Problems

For the following exercises, answer by proof, counterexample, or explanation.

For the following exercises, set up and evaluate each optimization problem.

For the following exercises, consider the construction of a pen to enclose an area.

For the following problems, consider a lifeguard at a circular pool with diameter $40\ \text{m}.$ He must reach someone who is drowning on the exact opposite side of the pool, at position $C.$ The lifeguard swims with a speed $v$ and runs around the pool at speed $w=3v.$

For the following exercises, consider a limousine that gets $m(v)=\frac{(120-2v)}{5}\ \text{mi/gal}$ at speed $v,$ the chauffeur costs $\text{\$15/h},$ and gas is $\text{\$}3.5\text{/}\text{gal}.$

For the following exercises, consider a pizzeria that sell pizzas for a revenue of $R(x)=ax$ and costs $C(x)=b+cx+d{x}^{2},$ where $x$ represents the number of pizzas $;a>c$.

For the following exercises, consider a wire $4\ \text{ft}$ long cut into two pieces. One piece forms a circle with radius $r$ and the other forms a square of side $x.$

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

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

A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides (). Given $100$ ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?

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Let $x$ denote the length of the side of the garden perpendicular to the rock wall and $y$ denote the length of the side parallel to the rock wall. Then the area of the garden is
\[A=x\cdot y.\]
We want to find the maximum possible area subject to the constraint that the total fencing is $100\ \text{ft}.$ From , the total amount of fencing used will be $2x+y.$ Therefore, the constraint equation is
\[2x+y=100.\]
Solving this equation for $y,$ we have $y=100-2x.$ Thus, we can write the area as
\[A(x)=x\cdot (100-2x)=100x-2{x}^{2}.\]
Before trying to maximize the area function $A(x)=100x-2{x}^{2},$ we need to determine the domain under consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need $x>0$ and $y>0.$ Since $y=100-2x,$ if $y>0,$ then $x<50.$ Therefore, we are trying to determine the maximum value of $A(x)$ for $x$ over the open interval $(0,50).$ We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval. Therefore, let’s consider the function $A(x)=100x-2{x}^{2}$ over the closed interval $[0,50].$ If the maximum value occurs at an interior point, then we have found the value $x$ in the open interval $(0,50)$ that maximizes the area of the garden. Therefore, we consider the following problem:
Maximize $A(x)=100x-2{x}^{2}$ over the interval $[0,50].$
As mentioned earlier, since $A$ is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, $A(x)=0.$ Since the area is positive for all $x$ in the open interval $(0,50),$ the maximum must occur at a critical point. Differentiating the function $A(x),$ we obtain
\[{A}^{'}(x)=100-4x.\]
Therefore, the only critical point is $x=25$ (). We conclude that the maximum area must occur when $x=25.$ Then we have $y=100-2x=100-2(25)=50.$ To maximize the area of the garden, let $x=25$ ft and $y=50\ \text{ft}.$ The area of this garden is $1250{\ \text{ft}}^{2}.$
Determine the maximum area if we want to make the same rectangular garden as in , but we have $200$ ft of fencing.

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The maximum area is $5000{\ \text{ft}}^{2}.$
An open-top box is to be made from a $24$ in. by $36$ in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?

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Step 1: Let $x$ be the side length of the square to be removed from each corner (). Then, the remaining four flaps can be folded up to form an open-top box. Let $V$ be the volume of the resulting box.
Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize $V.$
Step 3: As mentioned in step $2,$ we are trying to maximize the volume of a box. The volume of a box is $V=L\cdot W\cdot H,$ where $L,W,\ \text{and}\ H$ are the length, width, and height, respectively.
Step 4: From , we see that the height of the box is $x$ inches, the length is $36-2x$ inches, and the width is $24-2x$ inches. Therefore, the volume of the box is
\[V(x)=(36-2x)(24-2x)x=4{x}^{3}-120{x}^{2}+864x.\]
Step 5: To determine the domain of consideration, let’s examine . Certainly, we need $x>0.$ Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, $24$ in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for $x$ over the open interval $(0,12).$ Since $V$ is a continuous function over the closed interval $[0,12],$ we know $V$ will have an absolute maximum over the closed interval. Therefore, we consider $V$ over the closed interval $[0,12]$ and check whether the absolute maximum occurs at an interior point.
Step 6: Since $V(x)$ is a continuous function over the closed, bounded interval $[0,12],$ $V$ must have an absolute maximum (and an absolute minimum). Since $V(x)=0$ at the endpoints and $V(x)>0$ for $0\[{V}^{'}(x)=12{x}^{2}-240x+864.\]
To find the critical points, we need to solve the equation
\[12{x}^{2}-240x+864=0.\]
Dividing both sides of this equation by \(12,$ the problem simplifies to solving the equation
\[{x}^{2}-20x+72=0.\]
Using the quadratic formula, we find that the critical points are
\[x=\frac{20\text{\pm }\sqrt{{(-20)}^{2}-4(1)(72)}}{2}=\frac{20\text{\pm }\sqrt{112}}{2}=\frac{20\text{\pm }4\sqrt{7}}{2}=10\text{\pm }2\sqrt{7}.\]
Since $10+2\sqrt{7}$ is not in the domain of consideration, the only critical point we need to consider is $10-2\sqrt{7}.$ Therefore, the volume is maximized if we let $x=10-2\sqrt{7}\ \text{in}.$ The maximum volume is $V(10-2\sqrt{7})=640+448\sqrt{7}\approx 1825\ \text{in}{.}^{3}$ as shown in the following graph.
Suppose the dimensions of the cardboard in are 20 in. by 30 in. Let $x$ be the side length of each square and write the volume of the open-top box as a function of $x.$ Determine the domain of consideration for $x.$

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$V(x)=x(20-2x)(30-2x).$ The domain is $[0,10].$
An island is $2\ \text{mi}$ due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is $6\ \text{mi}$ west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of $8\ \text{mph}$ and swims at a rate of $3\ \text{mph}.$ How far should the visitor run before swimming to minimize the time it takes to reach the island?

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Step 1: Let $x$ be the distance running and let $y$ be the distance swimming (). Let $T$ be the time it takes to get from the cabin to the island.
Step 2: The problem is to minimize $T.$
Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance $=$ Rate $\ \times \$ Time $(D=R\ \times \ T),$ the time spent running is
\[{T}_{\text{running}}=\frac{{D}_{\text{running}}}{{R}_{\text{running}}}=\frac{x}{8},\]
and the time spent swimming is
\[{T}_{\text{swimming}}=\frac{{D}_{\text{swimming}}}{{R}_{\text{swimming}}}=\frac{y}{3}.\]
Therefore, the total time spent traveling is
\[T=\frac{x}{8}+\frac{y}{3}.\]
Step 4: From , the line segment of $y$ miles forms the hypotenuse of a right triangle with legs of length $2\ \text{mi}$ and $6-x\ \text{mi}.$ Therefore, by the Pythagorean theorem, ${2}^{2}+{(6-x)}^{2}={y}^{2},$ and we obtain $y=\sqrt{{(6-x)}^{2}+4}.$ Thus, the total time spent traveling is given by the function
\[T(x)=\frac{x}{8}+\frac{\sqrt{{(6-x)}^{2}+4}}{3}.\]
Step 5: From , we see that $0\le x\le 6.$ Therefore, $[0,6]$ is the domain of consideration.
Step 6: Since $T(x)$ is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let’s begin by looking for any critical points of $T$ over the interval $[0,6].$ The derivative is
\[{T}^{'}(x)=\frac{1}{8}-\frac{1}{2}\ \frac{{[{(6-x)}^{2}+4]}^{-1\text{/}2}}{3}\cdot 2(6-x)=\frac{1}{8}-\frac{(6-x)}{3\sqrt{{(6-x)}^{2}+4}}.\]
If ${T}^{'}(x)=0,$ then
\[\frac{1}{8}=\frac{6-x}{3\sqrt{{(6-x)}^{2}+4}}.\]
Therefore,
\[3\sqrt{{(6-x)}^{2}+4}=8(6-x).\]
Squaring both sides of this equation, we see that if $x$ satisfies this equation, then $x$ must satisfy
\[9[{(6-x)}^{2}+4]=64{(6-x)}^{2},\]
which implies
\[55{(6-x)}^{2}=36.\]
We conclude that if $x$ is a critical point, then $x$ satisfies
\[{(x-6)}^{2}=\frac{36}{55}.\]
Therefore, the possibilities for critical points are
\[x=6\text{\pm }\frac{6}{\sqrt{55}}.\]
Since $x=6+6\text{/}\sqrt{55}$ is not in the domain, it is not a possibility for a critical point. On the other hand, $x=6-6\text{/}\sqrt{55}$ is in the domain. Since we squared both sides of to arrive at the possible critical points, it remains to verify that $x=6-6\text{/}\sqrt{55}$ satisfies . Since $x=6-6\text{/}\sqrt{55}$ does satisfy that equation, we conclude that $x=6-6\text{/}\sqrt{55}$ is a critical point, and it is the only one. To justify that the time is minimized for this value of $x,$ we just need to check the values of $T(x)$ at the endpoints $x=0$ and $x=6,$ and compare them with the value of $T(x)$ at the critical point $x=6-6\text{/}\sqrt{55}.$ We find that $T(0)\approx 2.108\ \text{h}$ and $T(6)\approx 1.417\ \text{h,}$ whereas $T(6-6\text{/}\sqrt{55})\approx 1.368\ \text{h}.$ Therefore, we conclude that $T$ has a local minimum at $x\approx 5.19$ mi.
Suppose the island is $1$ mi from shore, and the distance from the cabin to the point on the shore closest to the island is $15\ \text{mi}.$ Suppose a visitor swims at the rate of $2.5\ \text{mph}$ and runs at a rate of $6\ \text{mph}.$ Let $x$ denote the distance the visitor will run before swimming, and find a function for the time it takes the visitor to get from the cabin to the island.

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$T(x)=\frac{x}{6}+\frac{\sqrt{{(15-x)}^{2}+1}}{2.5}$
Owners of a car rental company have determined that if they charge customers $p$ dollars per day to rent a car, where $50\le p\le 200,$ the number of cars $n$ they rent per day can be modeled by the linear function $n(p)=1000-5p.$ If they charge $\text{\$}50$ per day or less, they will rent all their cars. If they charge $\text{\$}200$ per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and $\text{\$}200$ per day to rent a car, how much should they charge to maximize their revenue?

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Step 1: Let $p$ be the price charged per car per day and let $n$ be the number of cars rented per day. Let $R$ be the revenue per day.
Step 2: The problem is to maximize $R.$
Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day—that is, $R=n\ \times \ p.$
Step 4: Since the number of cars rented per day is modeled by the linear function $n(p)=1000-5p,$ the revenue $R$ can be represented by the function
\[R(p)=n\ \times \ p=(1000-5p)p=-5{p}^{2}+1000p.\]
Step 5: Since the owners plan to charge between $\text{\$}50$ per car per day and $\text{\$}200$ per car per day, the problem is to find the maximum revenue $R(p)$ for $p$ in the closed interval $[50,200].$
Step 6: Since $R$ is a continuous function over the closed, bounded interval $[50,200],$ it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is ${R}^{'}(p)=-10p+1000.$ Therefore, the critical point is $p=100$ When $p=100,$ $R(100)=\text{\$}50,000.$ When $p=50,$ $R(p)=\text{\$}37,500.$ When $p=200,$ $R(p)=\text{\$}0.$ Therefore, the absolute maximum occurs at $p=\text{\$}100.$ The car rental company should charge $\text{\$}100$ per day per car to maximize revenue as shown in the following figure.
A car rental company charges its customers $p$ dollars per day, where $60\le p\le 150.$ It has found that the number of cars rented per day can be modeled by the linear function $n(p)=750-5p.$ How much should the company charge each customer to maximize revenue?

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The company should charge $\text{\$}75$ per car per day.
A rectangle is to be inscribed in the ellipse
\[\frac{{x}^{2}}{4}+{y}^{2}=1.\]
What should the dimensions of the rectangle be to maximize its area? What is the maximum area?

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Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let $L$ be the length of the rectangle and $W$ be its width. Let $A$ be the area of the rectangle.
Step 2: The problem is to maximize $A.$
Step 3: The area of the rectangle is $A=LW.$
Step 4: Let $(x,y)$ be the corner of the rectangle that lies in the first quadrant, as shown in . We can write length $L=2x$ and width $W=2y.$ Since $\frac{{x}^{2}}{4}+{y}^{2}=1$ and $y>0,$ we have $y=\sqrt{1-\frac{{x}^{2}}{4}}.$ Therefore, the area is
\[A=LW=(2x)(2y)=4x\sqrt{1-\frac{{x}^{2}}{4}}=2x\sqrt{4-{x}^{2}}.\]
Step 5: From , we see that to inscribe a rectangle in the ellipse, the $x$-coordinate of the corner in the first quadrant must satisfy $0
Step 6: As mentioned earlier, \(A(x)$ is a continuous function over the closed, bounded interval $[0,2].$ Therefore, it has an absolute maximum (and absolute minimum). At the endpoints $x=0$ and $x=2,$ $A(x)=0.$ For $00.$ Therefore, the maximum must occur at a critical point. Taking the derivative of $A(x),$ we obtain
\[\begin{array}{ll}A'(x) & =2\sqrt{4-{x}^{2}}+2x\cdot \frac{1}{2\sqrt{4-{x}^{2}}}(-2x) \\ & =2\sqrt{4-{x}^{2}}-\frac{2{x}^{2}}{\sqrt{4-{x}^{2}}} \\ & =\frac{8-4{x}^{2}}{\sqrt{4-{x}^{2}}}.\end{array}\]
To find critical points, we need to find where $A'(x)=0.$ We can see that if $x$ is a solution of
\[\frac{8-4{x}^{2}}{\sqrt{4-{x}^{2}}}=0,\]
then $x$ must satisfy
\[8-4{x}^{2}=0.\]
Therefore, ${x}^{2}=2.$ Thus, $x=\text{\pm }\sqrt{2}$ are the possible solutions of . Since we are considering $x$ over the interval $[0,2],$ $x=\sqrt{2}$ is a possibility for a critical point, but $x=\text{-}\sqrt{2}$ is not. Therefore, we check whether $\sqrt{2}$ is a solution of . Since $x=\sqrt{2}$ is a solution of , we conclude that $\sqrt{2}$ is the only critical point of $A(x)$ in the interval $[0,2].$ Therefore, $A(x)$ must have an absolute maximum at the critical point $x=\sqrt{2}.$ To determine the dimensions of the rectangle, we need to find the length $L$ and the width $W.$ If $x=\sqrt{2}$ then
\[y=\sqrt{1-\frac{{(\sqrt{2})}^{2}}{4}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}.\]
Therefore, the dimensions of the rectangle are $L=2x=2\sqrt{2}$ and $W=2y=\frac{2}{\sqrt{2}}=\sqrt{2}.$ The area of this rectangle is $A=LW=(2\sqrt{2})(\sqrt{2})=4.$
Modify the area function $A$ if the rectangle is to be inscribed in the unit circle ${x}^{2}+{y}^{2}=1.$ What is the domain of consideration?

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$A(x)=4x\sqrt{1-{x}^{2}}.$ The domain of consideration is $[0,1].$
A rectangular box with a square base, an open top, and a volume of $216$ in.³ is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?

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Step 1: Draw a rectangular box and introduce the variable $x$ to represent the length of each side of the square base; let $y$ represent the height of the box. Let $S$ denote the surface area of the open-top box.
Step 2: We need to minimize the surface area. Therefore, we need to minimize $S.$
Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is $x\cdot y.$ The area of the base is ${x}^{2}.$ Therefore, the surface area of the box is
\[S=4xy+{x}^{2}.\]
Step 4: Since the volume of this box is ${x}^{2}y$ and the volume is given as $216\ \text{in}{.}^{3},$ the constraint equation is
\[{x}^{2}y=216.\]
Solving the constraint equation for $y,$ we have $y=\frac{216}{{x}^{2}}.$ Therefore, we can write the surface area as a function of $x$ only:
\[S(x)=4x(\frac{216}{{x}^{2}})+{x}^{2}.\]
Therefore, $S(x)=\frac{864}{x}+{x}^{2}.$
Step 5: Since we are requiring that ${x}^{2}y=216,$ we cannot have $x=0.$ Therefore, we need $x>0.$ On the other hand, $x$ is allowed to have any positive value. Note that as $x$ becomes large, the height of the box $y$ becomes correspondingly small so that ${x}^{2}y=216.$ Similarly, as $x$ becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval $(0,\infty ).$ Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval $(0,\infty ).$
Step 6: Note that as $x\to {0}^{+},$ $S(x)\to \infty .$ Also, as $x\to \infty ,$ $S(x)\to \infty .$ Since $S$ is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some $x\in (0,\infty ).$ This minimum must occur at a critical point of $S.$ The derivative is
\[{S}^{'}(x)=-\frac{864}{{x}^{2}}+2x.\]
Therefore, ${S}^{'}(x)=0$ when $2x=\frac{864}{{x}^{2}}.$ Solving this equation for $x,$ we obtain ${x}^{3}=432,$ so $x=\sqrt[3]{432}=6\sqrt[3]{2}.$ Since this is the only critical point of $S,$ the absolute minimum must occur at $x=6\sqrt[3]{2}$ (see ). When $x=6\sqrt[3]{2},$ $y=\frac{216}{{(6\sqrt[3]{2})}^{2}}=3\sqrt[3]{2}\ \text{in}.$ Therefore, the dimensions of the box should be $x=6\sqrt[3]{2}\ \text{in}.$ and $y=3\sqrt[3]{2}\ \text{in}.$ With these dimensions, the surface area is
\[S(6\sqrt[3]{2})=\frac{864}{6\sqrt[3]{2}}+{(6\sqrt[3]{2})}^{2}=108\sqrt[3]{4}\ \text{in}{.}^{2}\]
Consider the same open-top box, which is to have volume $216\ \text{in}{.}^{3}.$ Suppose the cost of the material for the base is $20¢\text{/}\text{in}{.}^{2}$ and the cost of the material for the sides is $30¢\text{/}\text{in}{.}^{2}$ and we are trying to minimize the cost of this box. Write the cost as a function of the side lengths of the base. (Let $x$ be the side length of the base and $y$ be the height of the box.)

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$c(x)=\frac{259.2}{x}+0.2{x}^{2}$ dollars
When you find the maximum for an optimization problem, why do you need to check the sign of the derivative around the critical points?

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The critical points can be the minima, maxima, or neither.
Why do you need to check the endpoints for optimization problems?
True or False. For every continuous nonlinear function, you can find the value $x$ that maximizes the function.

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False; $y={x}^{2}$ has a minimum only
True or False. For every continuous nonconstant function on a closed, finite domain, there exists at least one $x$ that minimizes or maximizes the function.
To carry a suitcase on an airplane, the length $+\text{width}+$ height of the box must be less than or equal to $62\ \text{in}.$ Assuming the base of the suitcase is square, show that the volume is $V=h{(31-(\frac{1}{2})\ h)}^{2}.$ What height allows you to have the largest volume?

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$h=\frac{62}{3}$ in.
You are constructing a cardboard box with the dimensions $\text{2 m by 4 m}.$ You then cut equal-size squares from each corner so you may fold the edges. What are the dimensions of the box with the largest volume?
Find the positive integer that minimizes the sum of the number and its reciprocal.

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$1$
Find the two positive integers so that their sum is $10$ and the sum of their squares is as large as possible. Then find another two positive integers so that their sum is 10 and the sum of their squares is as small as possible.
You have $400\ \text{ft}$ of fencing to construct a rectangular pen for cattle. What are the dimensions of the pen that maximize the area?

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$100\ \text{ft by}\ 100\ \text{ft}$
You have $800\ \text{ft}$ of fencing to make a pen for hogs. If you have a river on one side of your property, what is the dimension of the rectangular pen that maximizes the area?
You need to construct a fence around an area of $1600\ {\text{ft}}^{2}.$ What are the dimensions of the rectangular pen to minimize the amount of material needed?

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$40\ \text{ft by}\ 40\ \text{ft}$
Two poles are connected by a wire that is also connected to the ground. The first pole is $20\ \text{ft}$ tall and the second pole is $10\ \text{ft}$ tall. There is a distance of $30\ \text{ft}$ between the two poles. Where should the wire be anchored to the ground to minimize the amount of wire needed?
[T] You are moving into a new apartment and notice there is a corner where the hallway narrows from $\text{8 ft to 6 ft}.$ What is the length of the longest item that can be carried horizontally around the corner?

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$19.73\ \text{ft}.$
A patient’s pulse measures $\text{70 bpm, 80 bpm, then 120 bpm}.$ To determine an accurate measurement of pulse, the doctor wants to know what value minimizes the expression ${(x-70)}^{2}+{(x-80)}^{2}+{(x-120)}^{2}?$ What value minimizes it?
In the previous problem, assume the patient was nervous during the third measurement, so we only weight that value half as much as the others. What is the value that minimizes ${(x-70)}^{2}+{(x-80)}^{2}+\frac{1}{2}{(x-120)}^{2}?$

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$84\ \text{bpm}$
You can run at a speed of $6$ mph and swim at a speed of $3$ mph and are located on the shore, $4$ miles east of an island that is $1$ mile north of the shoreline. How far should you run west to minimize the time needed to reach the island?
Find a function that measures the total amount of time it takes to reach the drowning person as a function of the swim angle, $\theta .$

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$T(\theta )=\frac{40\theta }{3v}+\frac{40\ \text{cos}\ \theta }{v}$
Find at what angle $\theta$ the lifeguard should swim to reach the drowning person in the least amount of time.
A truck uses gas as $g(v)=av+\frac{b}{v},$ where $v$ represents the speed of the truck and $g$ represents the gallons of fuel per mile. Assuming $a$ and $b$ are positive, at what speed is fuel consumption minimized?

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$v=\sqrt{\frac{b}{a}}$
Find the cost per mile at speed $v.$
Find the cheapest driving speed.

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approximately $34\ \text{mph}$
Find the profit function for the number of pizzas. How many pizzas gives the largest profit per pizza?
Assume that $R(x)=10x$ and $C(x)=2x+{x}^{2}.$ How many pizzas sold maximizes the profit?

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$4$
Assume that $R(x)=15x,$ and $C(x)=60+3x+\frac{1}{2}{x}^{2}.$ How many pizzas sold maximizes the profit?
Choose $x$ to maximize the sum of their areas.

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$0$
Choose $x$ to minimize the sum of their areas.
${x}^{2}{y}^{2}$
$y-\frac{1}{x}$

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Maximal: $x=1,y=9;$ minimal: none

Symbols used here

\infty

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

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\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.

\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 Problems

Set up and solve optimization problems in several applied fields.
Introduce all variables. If applicable, draw a figure and label all variables.
Determine which quantity is to be maximized or minimized, and for what range of values of the other variables (if this can be determined at this time).
Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more than one variable.
Write any equations relating the independent variables in the formula from step
Identify the domain of consideration for the function in step
Locate the maximum or minimum value of the function from step
To solve an optimization problem, begin by drawing a picture and introducing variables.

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.

Cubalah sendiri

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Lebih dalam 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