how to solve optimization problems calculus

$A drawing of a garden has x and y written on the vertical and horizontal sides, respectively. There is a rock wall running along the entire bottom horizontal length of the drawing.$

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

$A=x⋅y.$

We want to find the maximum possible area subject to the constraint that the total fencing is $100\,\text{ft}$. From Figure $\PageIndex{1}$, 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

$A(x)=x⋅(100−2x)=100x−2x^2.$

Before trying to maximize the area function $A(x)=100x−2x^2,$ 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 $A(x)=100x−2x^2$ 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 $A(x)=100x−2x^2$ 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 $x=25$ (Figure $\PageIndex{2}$). We conclude that the maximum area must occur when $x=25$.

$The function A(x) = 100x – 2x is graphed. At its maximum there is an intersection of two dashed lines and text that reads “Maximum area is 1250 square feet when x = 25 feet.”$

Then we have $y=100−2x=100−2(25)=50.$ To maximize the area of the garden, let $x=25\,\text{ft}$ and $y=50\,\text{ft}$. The area of this garden is $1250\, \text{ft}^2$.

Exercise $\PageIndex{1}$

Determine the maximum area if we want to make the same rectangular garden as in Figure $\PageIndex{2}$, but we have $200\,\text{ft}$ of fencing.

We need to maximize the function $A(x)=200x−2x^2$ over the interval $[0,100].$

The maximum area is $5000\, \text{ft}^2$.

Now let’s look at a general strategy for solving optimization problems similar to Example $\PageIndex{1}$.

Problem-Solving Strategy: Solving Optimization Problems

Now let’s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.

Example $\PageIndex{2}$: Maximizing the Volume of a Box

An open-top box is to be made from a $24\,\text{in.}$ by $36\,\text{in.}$ 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?

Step 1: Let $x$ be the side length of the square to be removed from each corner (Figure $\PageIndex{3}$). Then, the remaining four flaps can be folded up to form an open-top box. Let $V$ be the volume of the resulting box.

$There are two figures for this figure. The first one is a rectangle with sides 24 in and 36 in, with each corner having a square of side length x taken out of it. In the second picture, there is a box with side lengths x in, 24 – 2x in, and 36 – 2x in.$

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, are trying to maximize the volume of a box. The volume of a box is

\[V=L⋅W⋅H \nonumber, \nonumber \]

where $L,\,W,$and $H$ are the length, width, and height, respectively.

Step 4: From Figure $\PageIndex{3}$, 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

\[ \begin{align*} V(x) &=(36−2x)(24−2x)x \\[4pt] &=4x^3−120x^2+864x \end{align*}. \nonumber \]

Step 5: To determine the domain of consideration, let’s examine Figure $\PageIndex{3}$. 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, $24\,\text{in.}$; 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 $0<x<12,$ the maximum must occur at a critical point. The derivative is

$V′(x)=12x^2−240x+864.$

To find the critical points, we need to solve the equation

$12x^2−240x+864=0.$

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

\[\begin{align*} x &=\dfrac{20±\sqrt{(−20)^2−4(1)(72)}}{2} \\[4pt] &=\dfrac{20±\sqrt{112}}{2} \\[4pt] &=\dfrac{20±4\sqrt{7}}{2} \\[4pt] &=10±2\sqrt{7} \end{align*}. \nonumber \]

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

\[V(10−2\sqrt{7})=640+448\sqrt{7}≈1825\,\text{in}^3. \nonumber \]

as shown in the following graph.

$The function V(x) = 4x3 – 120x2 + 864x is graphed. At its maximum there is an intersection of two dashed lines and text that reads “Maximum volume is approximately 1825 cubic inches when x ≈ 4.7 inches.”$

Exercise $\PageIndex{2}$

Suppose the dimensions of the cardboard in Example $\PageIndex{2}$ are $20\,\text{in.}$ by $30\,\text{in.}$ 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 $x$.

The volume of the box is $L⋅W⋅H.$

$V(x)=x(20−2x)(30−2x).$ The domain is $[0,10]$.

Example $\PageIndex{3}$: Minimizing Travel Time

An island is $2$ mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is $6$ 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 $8$ mph and swims at a rate of $3$ mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?

Step 1: Let $x$ be the distance running and let $y$ be the distance swimming (Figure $\PageIndex{5}$). Let $T$ be the time it takes to get from the cabin to the island.

$The cabin is x miles from the shore. From that point on the shore, the island is y miles away. If you were to continue the line from the cabin to the shore (the x miles one) and if you were to draw a line from the island parallel to the shore, then the lines would extend 2 miles from the island and 6 miles from the cabin before intersecting.$

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 × Time $(D=R×T),$ the time spent running is

$T_{running}=\dfrac{D_{running}}{R_{running}}=\dfrac{x}{8}$,

and the time spent swimming is

$T_{swimming}=\dfrac{D_{swimming}}{R_{swimming}}=\dfrac{y}{3}$.

Therefore, the total time spent traveling is

$T=\dfrac{x}{8}+\dfrac{y}{3}$.

Step 4: From Figure $\PageIndex{5}$, the line segment of $y$ miles forms the hypotenuse of a right triangle with legs of length $2$ mi and $6−x$ 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

$T(x)=\dfrac{x}{8}+\dfrac{\sqrt{(6−x)^2+4}}{3}$.

Step 5: From Figure $\PageIndex{5}$, we see that $0≤x≤6$. 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

\[\begin{align*} T′(x) &=\dfrac{1}{8}−\dfrac{1}{2}\dfrac{[(6−x)^2+4]^{−1/2}}{3}⋅2(6−x) \\[4pt] &=\dfrac{1}{8}−\dfrac{(6−x)}{3\sqrt{(6−x)^2+4}} \end{align*}\]

If $T′(x)=0,$, then

\[\dfrac{1}{8}=\dfrac{6−x}{3\sqrt{(6−x)^2+4}} \label{ex3eq1} \]

\[3\sqrt{(6−x)^2+4}=8(6−x). \label{ex3eq2} \]

Squaring both sides of this equation, we see that if $x$ satisfies this equation, then $x$ must satisfy

\[9[(6−x)^2+4]=64(6−x)^2,\nonumber \]

which implies

\[55(6−x)^2=36. \nonumber \]

We conclude that if $x$ is a critical point, then $x$ satisfies

\[(x−6)^2=\dfrac{36}{55}. \nonumber \]

[Note that since we are squaring, $ (x-6)^2 = (6-x)^2.$]

Therefore, the possibilities for critical points are

\[x=6±\dfrac{6}{\sqrt{55}}.\nonumber \]

Since $x=6+6/\sqrt{55}$ is not in the domain, it is not a possibility for a critical point. On the other hand, $x=6−6/\sqrt{55}$ is in the domain. Since we squared both sides of Equation \ref{ex3eq2} to arrive at the possible critical points, it remains to verify that $x=6−6/\sqrt{55}$ satisfies Equation \ref{ex3eq1}. Since $x=6−6/\sqrt{55}$ does satisfy that equation, we conclude that $x=6−6/\sqrt{55}$ 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 $x=6−6/\sqrt{55}$. We find that $T(0)≈2.108\,\text{h}$ and $T(6)≈1.417\,\text{h}$, whereas

\[T(6−6/\sqrt{55})≈1.368\,\text{h}. \nonumber \]

Therefore, we conclude that $T$ has a local minimum at $x≈5.19$ mi.

Exercise $\PageIndex{3}$

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 $15$ mi. Suppose a visitor swims at the rate of $2.5$ mph and runs at a rate of $6$ 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.

The time $T=T_{running}+T_{swimming}.$

$T(x)=\dfrac{x}{6}+\dfrac{\sqrt{(15−x)^2+1}}{2.5} $

In business, companies are interested in maximizing revenue. In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let’s use these data to determine the price the company should charge to maximize the amount of money it brings in.

Example $\PageIndex{4}$: Maximizing Revenue

Owners of a car rental company have determined that if they charge customers $p$ dollars per day to rent a car, where $50≤p≤200$, the number of cars $n$ they rent per day can be modeled by the linear function $n(p)=1000−5p$. If they charge $$50$ per day or less, they will rent all their cars. If they charge $$200$ per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $$50$ per day and $$200$ per day to rent a car, how much should they charge to maximize their revenue?

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, $R=n×p.$

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

\[ \begin{align*} R(p) &=n×p \\[4pt] &=(1000−5p)p \\[4pt] &=−5p^2+1000p.\end{align*}\]

Step 5: Since the owners plan to charge between $$50$ per car per day and $$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)=$50,000.$ When $p=50, R(p)=$37,500$. When $p=200, R(p)=$0$.

Therefore, the absolute maximum occurs at $p=$100$. The car rental company should charge $$100$ per day per car to maximize revenue as shown in the following figure.

$The function R(p) is graphed. At its maximum there is an intersection of two dashed lines and text that reads “Maximum revenue is $50,000 per day when the price charged per car is $100 per day.”$

Exercise $\PageIndex{4}$

A car rental company charges its customers $p$ dollars per day, where $60≤p≤150$. 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?

$R(p)=n×p,$ where $n$ is the number of cars rented and $p$ is the price charged per car.

The company should charge $$75$ per car per day.

Example $\PageIndex{5}$: Maximizing the Area of an Inscribed Rectangle

A rectangle is to be inscribed in the ellipse

\[\dfrac{x^2}{4}+y^2=1. \nonumber \]

What should the dimensions of the rectangle be to maximize its area? What is the maximum area?

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.

$The ellipse x2/4 + y2 = 1 is drawn with its x intercepts being ±2 and its y intercepts being ±1. There is a rectangle inscribed in the ellipse with length L (in the x-direction) and width W.$

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 Figure $\PageIndex{7}$. We can write length $L=2x$ and width $W=2y$. Since $\dfrac{x^2}{4}+y^2=1$ and $y>0$, we have $y=\sqrt{1-\dfrac{x^2}{4}}$. Therefore, the area is

$A=LW=(2x)(2y)=4x\sqrt{1-\dfrac{x^2}{4}}=2x\sqrt{4−x^2}$

Step 5: From Figure $\PageIndex{7}$, we see that to inscribe a rectangle in the ellipse, the $x$-coordinate of the corner in the first quadrant must satisfy $0<x<2$. Therefore, the problem reduces to looking for the maximum value of $A(x)$ over the open interval $(0,2)$. Since $A(x)$ will have an absolute maximum (and absolute minimum) over the closed interval $[0,2]$, we consider $A(x)=2x\sqrt{4−x^2}$ over the interval $[0,2]$. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.

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 $0<x<2$, $A(x)>0$.

Therefore, the maximum must occur at a critical point. Taking the derivative of $A(x)$, we obtain

\[ \begin{align*} A'(x) &=2\sqrt{4−x^2}+2x⋅\dfrac{1}{2\sqrt{4−x^2}}(−2x) \\[4pt] &=2\sqrt{4−x^2}−\dfrac{2x^2}{\sqrt{4−x^2}} \\[4pt] &=\dfrac{8−4x^2}{\sqrt{4−x^2}} . \end{align*}\]

To find critical points, we need to find where $A'(x)=0.$ We can see that if $x$ is a solution of

\[\dfrac{8−4x^2}{\sqrt{4−x^2}}=0, \label{ex5eq1} \]

then $x$ must satisfy

\[8−4x^2=0. \nonumber \]

Therefore, $x^2=2.$ Thus, $x=±\sqrt{2}$ are the possible solutions of Equation \ref{ex5eq1}. Since we are considering $x$ over the interval $[0,2]$, $x=\sqrt{2}$ is a possibility for a critical point, but $x=−\sqrt{2}$ is not. Therefore, we check whether $\sqrt{2}$ is a solution of Equation \ref{ex5eq1}. Since $x=\sqrt{2}$ is a solution of Equation \ref{ex5eq1}, 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

\[y=\sqrt{1−\dfrac{(\sqrt{2})^2}{4}}=\sqrt{1−\dfrac{1}{2}}=\dfrac{1}{\sqrt{2}}.\nonumber \]

Therefore, the dimensions of the rectangle are $L=2x=2\sqrt{2}$ and $W=2y=\dfrac{2}{\sqrt{2}}=\sqrt{2}$. The area of this rectangle is $ A=LW=(2\sqrt{2})(\sqrt{2})=4.$

Exercise $\PageIndex{5}$

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?

If $(x,y)$ is the vertex of the square that lies in the first quadrant, then the area of the square is $A=(2x)(2y)=4xy.$

$A(x)=4x\sqrt{1−x^2}.$ The domain of consideration is $[0,1]$.

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 $(−∞,∞)$ 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 $(0,∞),$ 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.

Example $\PageIndex{6}$: Minimizing Surface Area

A rectangular box with a square base, an open top, and a volume of $216 \,\text{in}^3$ 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?

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.

$A box with square base is shown. The base has side length x, and the height is y.$

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 $x⋅y.$ 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 $x^2y$ and the volume is given as $216\,\text{in}^3$, the constraint equation is

$x^2y=216$.

Solving the constraint equation for $y$, we have $y=\dfrac{216}{x^2}$. Therefore, we can write the surface area as a function of $x$ only:

\[S(x)=4x\left(\dfrac{216}{x^2}\right)+x^2.\nonumber \]

Therefore, $S(x)=\dfrac{864}{x}+x^2$.

Step 5: Since we are requiring that $x^2y=216$, 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 $x^2y=216$. Similarly, as $x$ becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval $(0,∞)$. 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 $(0,∞).$

Step 6: Note that as $x→0^+,\, S(x)→∞.$ Also, as $x→∞, \,S(x)→∞$. Since $S$ is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some $x∈(0,∞)$. This minimum must occur at a critical point of $S$. The derivative is

\[S′(x)=−\dfrac{864}{x^2}+2x.\nonumber \]

Therefore, $S′(x)=0$ when $2x=\dfrac{864}{x^2}$. 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 Figure $\PageIndex{9}$).

When $x=6\sqrt[3]{2}$, $y=\dfrac{216}{(6\sqrt[3]{2})^2}=3\sqrt[3]{2}\,\text{in.}$ 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

\[S(6\sqrt[3]{2})=\dfrac{864}{6\sqrt[3]{2}}+(6\sqrt[3]{2})^2=108\sqrt[3]{4}\,\text{in}^2\nonumber \]

$The function S(x) = 864/x + x2 is graphed. At its minimum there is a dashed line and text that reads “Minimum surface area is 108 times the cube root of 4 square inches when x = 6 times the cube root of 2 inches.”$

Exercise $\PageIndex{6}$

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 $20¢/\text{in}^2$ and the cost of the material for the sides is $30¢/\text{in}^2$ 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.)

If the cost of one of the sides is $30¢/\text{in}^2,$ the cost of that side is $0.30xy$ dollars.

$c(x)=\dfrac{259.2}{x}+0.2x^2$ dollars

Key Concepts

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AP®︎/College Calculus AB

Unit 5: lesson 11.

Optimization: sum of squares

What dimensions should his enclosure be for the area to be maximized?

Step 2: Find, or identify, the constraint and write an expression for it in terms of the same variables you used in step 1 .

Note: there will always be a constraint. The constraint is what is stopping us from making the quantity $Q$ inifinitely large, or small. Every optimization problem has a constraint.

Step 3: Use the expression found for the constraint, in step 2 , to eliminate one of the variables in the expression for $Q$.

Note: at the end of this step, the expression for $Q$ should only have one variable; it is therefore a function of one variable only.

Step 4: use calculus to optimize , in other words: use calculus to find the maximum or minimum value of the expression (or the function) for $Q$ obtained at the end of step 3 .

Important Note: although it's quite clear that this value of $r$ can only correspond to a minimum surface area (since, in theory, without any constraints on the amount of material we can use: there is no limit to how big we can make the can and therefore no maximum surface area) if needs be: we can confirm that this value of $r$ does indeed correspond to a minimum surface area using the second derivative test .

To find the second derivative, $S''(r)$ we differentiate $S'(r) = 4\pi r - \frac{660}{r^2}$ with respect to $r$: \[S''(r) = 4 \pi + \frac{1320}{r^3}\] So when $r = \sqrt[3]{\frac{165}{\pi}}$ that's: \[S''\begin{pmatrix}\sqrt[3]{\frac{165}{ \pi}} \end{pmatrix} = 4\pi + \frac{1320}{\begin{pmatrix} \sqrt[3]{\frac{165}{ \pi}} \end{pmatrix}^3}\] That's: \[S''\begin{pmatrix}\sqrt[3]{\frac{165}{\pi}} \end{pmatrix} = 4\pi + \frac{1320}{\frac{165}{\pi}}\] Remembering that all that matters, when using the second derivative test is the sign or the second derivative , it's quite clear that $S''\begin{pmatrix}\sqrt[3]{\frac{165}{\pi}} \end{pmatrix}$ is positive (since both terms in the sum are positive). We can therefore state: \[S''\begin{pmatrix}\sqrt[3]{\frac{165}{\pi}} \end{pmatrix} > 0\] and that the second derivative test allows us to state that when $r = \sqrt[3]{\frac{165}{\pi}} \approx 3.74 \ \text{cm}$ the surface area, $S(r)$, reaches a minimum value .

The corresponding height, $h$, of the can can/should then be found using the expression we wrote at the beginning of Step 3 , $h = \frac{330}{\pi r^2}$, using $r = \sqrt[3]{\frac{165}{\pi}}$, that's: \[h = \frac{330}{\pi \begin{pmatrix} \sqrt[3]{\frac{165}{ \pi}}\end{pmatrix}^2} \approx 7.49 \ \text{cm}\]

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IMAGES

CA Calculus BC Video Lesson 3-7 Optimization Word Problems
Learn how to solve optimization problems and find the extremes, the local or global minima or
Calculus: Optimization Problems
Rectangular Fence Dimensions to Minimize Cost
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How to Solve Optimization Problems in Calculus
Stage I: Develop the function. · Draw a picture of the physical situation. · Write an equation that relates the quantity you want to optimize in
Optimization Problems
This calculus video explains how to solve optimization problems. It explains how to solve the fence along the river problem
How to Solve ANY Optimization Problem [Calc 1]
Optimization Problems - Calculus · Solving Optimization Problems in 5 Steps EXPLAINED with Examples · Optimization: profit | Applications of
Optimization In Calculus (How-To w/ 7 Step-by-Step Examples!)
Solving Optimization Problems (Step-by-Step) ... Step 1: Translate the problem using assign symbols, variables, and sketches, when applicable, by
Optimization
In optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to solve one kind of
4.7: Optimization Problems
Problem-Solving Strategy: Solving Optimization Problems · Introduce all variables. · Determine which quantity is to be maximized or minimized, and
5.8 Optimization Problems
Guideline for Solving Optimization Problems. · Identify what is to be maximized or minimized and what the constraints are. · Draw a diagram (if appropriate) and
Optimization: box volume (Part 1) (video)
How would I go about solving this circle optimization problem?
Steps for Optimization Problems 1. Draw and label a picture
4. Reduce the objective function to one variable. Solve each of the constraint equations for one of the variables and substitute this into the objective
Optimization Problems
How to solve classic optimization problems. Optimization Problems with calculus consist of maximizing, or minimizing, a quantity under a given constraint.
How to Solve an Optimization Problem by Finding a Minimum Value
Step 1: Draw a figure of your problem. Introduce variables and label what each variable represents. Step 2: Determine the value that is being minimized. Write a