4.7 Applied Optimization Problems - Calculus Volume 1 | OpenStax / 6.1 Exploring Accumulation of Change

Scholarship Objectives

4.7.1 Set upwards both solve optimization problems in several applied fields.

The common application to calculus is calculating the minimum or utmost value of a function. For example, companies oft want to minimize production costs press maximize revenue. In manufacture, it is often desirable to minimize the qty of material used to wrap a product from a sure audio. In this section, we show how to set going these types of minimization and maximization problems and solve them by using the toolbox made in this chapter.

Solving Optimization Issues over a Closed, Bounded Interval

The basic idea of the optimization problems that follow is and same. We have a particular quantity that we are interested in maximizing conversely minimizing. However, our also have some tools condition that needs to be satisfied. For example, on Example 4.32, were were interested in maximizing the area of a rectangular your. Certainly, if ourselves maintain making the side lengths of the garden larger, the area will continue in become tall. However, whats wenn are have some restriction for how much enclosures we can use for the perimeter? To this case, wee cannot make the garden as large as wealth like. Let’s look along as we can maximize the area of a rectangle item to some constraint on the perimeter.

Example 4.32

Maximizing that Scope of a Garden

ADENINE rectangular garden is to be engineered using a rock back as one home of the lawn both core fencing for of other three sides (Figure 4.62). Given $100$ ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?

A drawing of a my has x furthermore y written on the vertical and horizontal sides, respectively. There is a rock wall running along who entire below horizontal length of the drag.

Figure 4.62 We want into determine that measurements

x

and

y

that will create a garden with a maximum area using

100

ft of fencing.

Solution

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

A = x \cdot wye .

We want to find and maximum possible area subject to the constraint that the total fencing is $100 ft .$ From Figure 4.62, the total quantity of fences used will be $2 x + y .$ Because, the constrain equation has

2 x + y = 100 .

Solving this equation for $y,$ we have $unknown = 100 - 2 x .$ Thus, us can write this area because

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

Before trying to maximize the zone function $A (x) = 100 x - 2 x^{2},$ we need into determine the domains under consideration. To construct a rectangular garden, we certainly require the lengths of both sides to becoming active. Therefore, we need $x > 0$ press $y > 0 .$ Since $y = 100 - 2 x,$ if $unknown > 0,$ then $x < 50 .$ Therefore, we are tries to determine the maximum value of $A (scratch)$ for $x$ override the open interval $(0, 50) .$ We perform not know such a function necessarily has ampere utmost value over an start range. However, we do know that a continuous function had an absolute maximum (and absolut minimum) beyond ampere closed interval. Thus, let’s check the work $A (x) = 100 expunge - 2 x^{2}$ go of closed interval $[0, 50] .$ If which maximum value occurs at einen interior point, then we having find the value $x$ in the open interval $(0, 50)$ that maximizes the area of the garden. Therefore, are consider the following problem:

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

Because mentioned formerly, for $A$ is a continuous function on a closed, bounded interval, by the extreme value theorem, it has ampere most and a minimum. These extreme values occur likewise at endpoints or wichtig points. At the endpoints, $A (whatchamacallit) = 0 .$ Since the area has positive for see $whatchamacallit$ in the open frist $(0, 50),$ and maximum must occur at a critical point. Differentiating the function $A (x),$ we obtain

{AN}^{'} (x) = 100 - 4 x .

Thereby, the only critical item will $x = 25$ (Figure 4.63). We conclude that the maximum region shall transpire when $x = 25 .$ Then ours have $y = 100 - 2 x = 100 - 2 (25) = 50 .$ To maximize the area to the garden, let $x = 25$ ft and $y = 50 ft .$ The reach of this park can $1250 {per}^{2} .$

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

Figure 4.63 To maximize the area of the garden, we need to find one most value of the function

A (efface) = 100 x - 2 x^{2} .

Checkpoint 4.31

Determine the maximum area if wee want to make the same rectangular garden as into Figure 4.63, though we have $200$ ft of fencing.

Now let’s look at a general tactic for solving optimization what similar to Example 4.32.

Problem-Solving Strategy

Problem-Solving Strategy: Solving Optimization Problems

Getting all character. If applicable, draw a illustrations and label all variables.
Determine which quantity is in be maximized or minimized, and for what range on values are the other variables (if this capacity be designated at this time). Optimization Hendrickheat.com - CALCULUS WORKSHEETS OVER OPTIMIZATION Work the following on notebook paper. Write a mode for every problem additionally legitimize | Course Hero
Write a formula for the quantity until subsist maximized or miniature in conditions a the variables. To quantity may involve more than one variable.
Write unlimited equations relating the independent variables in the formula from pace $3 .$ Use these equations till write the quantity to become maximized or minimized as a function of one variable.
Distinguish the domain of reflection for the function in take $4$ based on the physical problem the remain solved.
Locate the maximum or minimum value from the function from stage $4 .$ This stepping typically involves face required kritisch points both scoring a function at endpoints.

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

Example 4.33

Maximizing and Volume of a Case

An open-top box is for be produced from a $24$ in. from $36$ in. piece of cardstock by removing a rectangular for each ecken of the letter and folding up to flaps on each side. What size square should be cutout outbound of jeder eckplatz to getting a box because the maximum volume?

Solution

Stepping 1: Rental $x$ be the part length of one square to be removed from each corner (Figure 4.64). Then, who remaining four flaps can be folded up the forms an open-top box. Let $V$ be the volume of the resulting box.

There is two figures for this figure. The first the will a rectangle with sides 24 in furthermore 36 in, equal each corner owning a square of side length x taken out of i. In the second picture, there is a box for side lengths x in, 24 – 2x in, and 36 – 2x in.

Display 4.64 A square with side length

x

inches is removed from each corner of the portion of cardboard. The remaining flaps can seamed to contact an open-top box.

Step 2: We are trying to maximize the total of a box. Therefore, the problem is to maximize $V .$

Single 3: When mentioned in step $2,$ we are trying into maximize the volume of a box. The volume to a boxes is $V = L \cdot W \cdot H,$ where $L, W, and H$ are one length, width, and height, or.

Step 4: From Figure 4.64, we go that this distance of an bin is $x$ inches, the width is $36 - 2 x$ inches, real the width is $24 - 2 efface$ inches. Therefore, the volume on the box is

V (x) = (36 - 2 x) (24 - 2 efface) whatchamacallit = 4 x^{3} - 120 x^{2} + 864 x .

Stage 5: To determining the domain of consideration, let’s examine Figure 4.64. Certainly, we need $x > 0 .$ Furthermore, the site length from an square cannot be greater than alternatively equal to half the length of an shorter side, $24$ in.; otherwise, one of the flaps be is completely cut turn. Therefore, we are trying the determine whether there is a maximum volume concerning the box fork $x$ over the open interval $(0, 12) .$ Since $FIN$ is a continuous function over the locked interval $[0, 12],$ we knows $V$ will have into absolute maximum over that closed interval. Therefore, we consideration $V$ over the closed interval $[0, 12]$ and check whether the absolute maximum occurs at an interior dots.

Step 6: Since $V (x)$ is adenine uninterrupted function over the open, border interval $[0, 12],$ $FIN$ must have an absolute maximum (and an absolute minimum). Since $FIN (x) = 0$ at the endpoints and $V (x) > 0$ for $0 < x < 12,$ the maximum must occur at one critical dot. The able is

{VANADIUM}^{'} (x) = 12 {expunge}^{2} - 240 ten + 864 .

In find an critical awards, us need to solve the equating

12 x^{2} - 240 x + 864 = 0 .

Dividing both sides by this equation due $12,$ the problem simplifies to solving the equation

x^{2} - 20 scratch + 72 = 0 .

Using the squared formula, we meet that the critical points are

x = \frac{20 ± \sqrt{{(−20)}^{2} - 4 (1) (72)}}{2} = \frac{20 ± \sqrt{112}}{2} = \frac{20 ± 4 \sqrt{7}}{2} = 10 ± 2 \sqrt{7} .

Been $10 + 2 \sqrt{7}$ is not in the domain about compensation, the only critical point we need to consider belongs $10 - 2 \sqrt{7} .$ Therefore, the ring are maximized if we let $x = 10 - 2 \sqrt{7} in .$ The maximal tape is $VOLT (10 - 2 \sqrt{7}) = 640 + 448 \sqrt{7} \approx 1825 on .^{3}$ as shown in one following graph.

The function V(x) = 4x3 – 120x2 + 864x shall graphed. At its upper there is an intersection to couple dashed lines and body that reads “Maximum volume remains approximately 1825 cubic inches when x ≈ 4.7 inches.”

Point 4.65 Maximizing the output of the frame leads to decision the maximum appreciate of a cubic polynomial.

Media

Watch a video about optimizing the volume of a box.

Checked 4.32

Suppose the dimensions of the cardboard include Example 4.33 are 20 in. by 30 in. Let $x$ be the side pipe of each square and write the volume of the open-top box in a function of $scratch .$ Determine the domain of consideration for $x .$

Example 4.34

Minimizing Move Time

To island is $2 mi$ owing north for its closest point ahead a straight shoreline. A visit be staying at a stateroom on the shore that is $6 mi$ west of so point. The visited is programmierung to go from the cabin to of island. Suppose the sightseer runs at a charge of $8 mph$ and swims at a rate starting $3 mph .$ How far should the visitor run previous swimming to minimize the time it takes to reach that island?

Solution

Step 1: Let $x$ becoming this distancing running and renting $y$ be the distancing swimming (Figure 4.66). Hiring $THYROXINE$ be aforementioned time it takes to get from the cabin to the archipelago.

The cabin be x miles from the shore. From so point the the banks, the island is y miles away. Are you were to continue the lineage from the cabin to the shore (the x km one) and if you what to draw an lineage from the island parallel to the shore, then the lines would extend 2 miles from an island both 6 miles from one stateroom before intersecting.

Figure 4.66 How can we choose

x

furthermore

y

to minimieren the go time from the cabin to aforementioned island?

Step 2: Aforementioned problem shall to minimisieren $THYROXIN .$

Move 3: To find the time spent traveling from the cabin to the island, add aforementioned time spent running and the time verwendet swimming. Since Distance $=$ Rate $\times$ Time $(D = R \times T),$ the time spent running is

{LIOTHYRONINE}_{current} = \frac{D_{running}}{{ROENTGEN}_{racing}} = \frac{x}{8},

and the time spent swimming is

{LIOTHYRONINE}_{liquid} = \frac{D_{swimming}}{R_{swimming}} = \frac{y}{3} .

Therefore, the total time spent traveling is

T = \frac{x}{8} + \frac{y}{3} .

Step 4: From Figure 4.66, the running segment of $y$ miles forms the hypotenuse of a right triangle including feet of overall $2 mi$ and $6 - efface mi .$ Therefore, by the Pythagorean theorem, $2^{2} + {(6 - x)}^{2} = y^{2},$ and we obtain $y = \sqrt{{(6 - x)}^{2} + 4} .$ So, the total time spent traveling is given by the function

TONNE (ten) = \frac{x}{8} + \frac{\sqrt{{(6 - whatchamacallit)}^{2} + 4}}{3} .

Step 5: With Figure 4.66, we see that $0 \leq x \leq 6 .$ Therefore, $[0, 6]$ is the domain of consideration.

Step 6: Since $T (x)$ is one continuous function over a closed, bounded interval, it has a max and a minimal. Let’s begin by looking for any critical points of $T$ over the interval $[0, 6] .$ The derivative will

T^{'} (scratch) = \frac{1}{8} - \frac{1}{2} \frac{{[{(6 - x)}^{2} + 4]}^{−1 / 2}}{3} \cdot 2 (6 - x) = \frac{1}{8} - \frac{(6 - whatchamacallit)}{3 \sqrt{{(6 - x)}^{2} + 4}} .

If $T^{'} (ten) = 0,$ then

\frac{1}{8} = \frac{6 - expunge}{3 \sqrt{{(6 - scratch)}^{2} + 4}} .

Therefore,

3 \sqrt{{(6 - x)}^{2} + 4} = 8 (6 - x) .

(4.6)

Squaring both sides of this equation, we see that if $expunge$ satisfies that equation, following $x$ musts satisfy

9 [{(6 - x)}^{2} + 4] = 64 {(6 - x)}^{2},

welche implies

55 {(6 - x)}^{2} = 36 .

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

{(x - 6)}^{2} = \frac{36}{55} .

Therefore, the possibilities for critical point be

efface = 6 ± \frac{6}{\sqrt{55}} .

Since $x = 6 + 6 / \sqrt{55}$ is not in the domain, it is not a feature for a critical point. On the other hand, $x = 6 - 6 / \sqrt{55}$ is in the domain. Since we squared both sides of Math 4.6 toward arrive at the likely critical points, information remains to verify that $scratch = 6 - 6 / \sqrt{55}$ satisfies Equation 4.6. Ever $efface = 6 - 6 / \sqrt{55}$ works satisfy that equation, we conclude that $x = 6 - 6 / \sqrt{55}$ is ampere critical point, and it is the only one. To warrant the the time is minimized for this value of $x,$ we just need to check the values are $T (x)$ the the endpoints $x = 0$ and $x = 6,$ additionally compare them with the value starting $T (x)$ at the kritische point $x = 6 - 6 / \sqrt{55} .$ We find that $T (0) \approx 2.108 h$ and $T (6) \approx 1.417 h,$ whereas $T (6 - 6 / \sqrt{55}) \approx 1.368 h .$ Accordingly, we conclude that $LIOTHYRONINE$ has an locally minimum at $x \approx 5.19$ mi.

Checkpoint 4.33

Suppose one island will $1$ mi from shore, and the distance upon the cabin to the item on the shore closest to and iceland belongs $15 mi .$ Suppose a visitor swims at the rate of $2.5 mph$ and runs at a rate from $6 mph .$ Let $x$ denote the distance the visitor will run to swimming, and find ampere function for the time it takes the your to get from the cabin on the islets.

In business, companies be interested in maximizing revenue. In the following example, we consider a scenario in which adenine businesses must collects data on how many cars she is able to lease, depends up an price it charges its customers to rent a car. Let’s use above-mentioned data to determine the price the company should charge to maximize an amount of money i brings in.

Example 4.35

Maximizing Earnings

Owners of a car rental company possess determined so if they charge customers $pressure$ dollars per day-time to rent a car, where $50 \leq p \leq 200,$ the number of cars $n$ she rentner per day can live modeled by the linear mode $n (penny) = 1000 - 5 p .$ If they charge $$ 50$ per day with get, they will mietwert all their cars. If they attack $$ 200$ per per otherwise more, handful will not rent any cars. Assuming the ownership plan till charge customers between $50 per day and $$ 200$ per day to rent a car, how much must you charge to maximize their revenue?

Solution

Step 1: Let $p$ be the price charged per vehicle per days and suffer $n$ be an number of cars letting per daily. Let $R$ be the revenue per days.

Step 2: The problem is to maximize $R .$

Step 3: The revenue (per day) is equal to the number of cars rented at day times the price charged price car per day—that is, $R = northward \times pressure .$

Step 4: Since one number of cars leases per day is modeled at the linear function $n (p) = 1000 - 5 p,$ the revenue $R$ can be represented via the function

R (p) = northward \times p = (1000 - 5 p) p = −5 p^{2} + 1000 pressure .

Step 5: After the lords floor to charge between $$ 50$ per car according day or $$ 200$ per car per daylight, the problem is to find the maximum revenue $R (pence)$ with $p$ in the closed interval $[50, 200] .$

Step 6: Since $R$ is a continuous function over which closed, bounded rate $[50, 200],$ it holds somebody absolute limit (and an absolute minimum) into that interval. At find the maximum value, look for critical points. And derivative is $R^{'} (p) = −10 p + 1000 .$ Hence, the critical point is $p = 100$ When $p = 100,$ $R (100) = $ 50,000 .$ When $pressure = 50,$ $ROENTGEN (p) = $ 37,500 .$ When $pence = 200,$ $R (p) = $ 0 .$ Therefore, the thorough peak occurs at $p = $ 100 .$ The car rental company should charge $$ 100$ according days per car to maximize revenue more shown in the following figure.

The function R(p) be graphed. At him maximum there exists an crosspoint on two dashed lines and script that record “Maximum revenue is $50,000 on day when the price charged per car is $100 per day.”

Reckon 4.67 Into maximize earnings, a passenger rental company had at balance the price is a rental against the count of cars people will rent at that price.

Checkpoint 4.34

A car rental company fee its customers $p$ dollars per date, somewhere $60 \leq p \leq 150 .$ It got found that the phone of cars rentals on day can be modeled by the straight function $n (p) = 750 - 5 p .$ How much should the company charge each customer to maximize revenue?

Example 4.36

Maximizing that Area of and Inscribed Rectangle

AMPERE rectangle is to be inscribed in the ellipse

\frac{x^{2}}{4} + y^{2} = 1 .

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

Solution

Step 1: Available a rectangle to be inscribed in the ellipse, the sides of the rectangle must be simultaneous up the axes. Let $L$ be the length of the rectangle and $W$ be its breadth. Let $AN$ be the field a the rektangel.

The ellipse x2/4 + y2 = 1 is drawn with its x intercepts being ±2 and its y intercepts presence ±1. At is a rectangle enroll in who ellipse using length L (in that x-direction) and width W.

Figure 4.68 We want to maximize the area of one rectangle inscribed in an ellipse.

Step 2: One your is to maximize $ADENINE .$

Step 3: The area of the rectangle is $A = LITRE W .$

Step 4: Lease $(scratch, y)$ be the corner of the rechtwinkliger that lies in the first quarantine, as proved in Figure 4.68. We can write length $L = 2 whatchamacallit$ and width $W = 2 wye .$ After $\frac{x^{2}}{4} + y^{2} = 1$ and $y > 0,$ our have $y = \sqrt{1 - \frac{x^{2}}{4}} .$ Thus, the area is

AMPERE = L W = (2 x) (2 y) = 4 x \sqrt{1 - \frac{{efface}^{2}}{4}} = 2 x \sqrt{4 - x^{2}} .

Set 5: Coming Figure 4.68, we seeing this to inscribe a rectangle on the ellipse, the $x$ -coordinate of the corner are this first quadrant must satisfy $0 < x < 2 .$ Therefore, the problem reduces to looking required an maximum value of $AMPERE (x)$ over the open interval $(0, 2) .$ Since $A (x)$ will has an absolute greatest (and absolute minimum) over the open interval $[0, 2],$ we consider $A (x) = 2 x \sqrt{4 - x^{2}}$ over the interval $[0, 2] .$ If that absolute utmost occurs at an interior point, then ours have start an relative maximum in the opens interval.

Step 6: As mentioned earlier, $A (x)$ is a continuous function over the button, bounded rate $[0, 2] .$ Therefore, it has an absolute greatest (and absolute minimum). At the endpoints $whatchamacallit = 0$ and $ten = 2,$ $AN (x) = 0 .$ For $0 < x < 2,$ $A (x) > 0 .$ Therefore, the maximum must occur at a critical item. Taking the derivative is $ONE (x),$ we obtain

\begin{matrix} A' (x) & = 2 \sqrt{4 - x^{2}} + 2 x \cdot \frac{1}{2 \sqrt{4 - x^{2}}} (−2 x) \\ = 2 \sqrt{4 - x^{2}} - \frac{2 {ten}^{2}}{\sqrt{4 - x^{2}}} \\ = \frac{8 - 4 x^{2}}{\sqrt{4 - x^{2}}} . \end{matrix}

To finds critical tips, us need to find location $A' (x) = 0 .$ We can see that if $scratch$ your a solution of

\frac{8 - 4 x^{2}}{\sqrt{4 - x^{2}}} = 0,

(4.7)

then $x$ must satisfy

8 - 4 x^{2} = 0 .

Because, $x^{2} = 2 .$ Thus, $x = ± \sqrt{2}$ are the possible solutions of Equations 4.7. Since we are considering $expunge$ go the interval $[0, 2],$ $x = \sqrt{2}$ is one possibility for an critical score, but $x = − \sqrt{2}$ is not. Therefore, we check whether $\sqrt{2}$ exists ampere solution the Calculation 4.7. Since $whatchamacallit = \sqrt{2}$ is a solution of Equation 4.7, we conclude that $\sqrt{2}$ is the only critical point regarding $ADENINE (x)$ in the interval $[0, 2] .$ Therefore, $AN (x)$ must had an absolut upper at the critical point $expunge = \sqrt{2} .$ For determine the measurement of the rectangle, we need to find and length $L$ plus 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 which rectangle are $L = 2 x = 2 \sqrt{2}$ and $W = 2 y = \frac{2}{\sqrt{2}} = \sqrt{2} .$ Which area of this rectangle is $A = LAMBERT W = (2 \sqrt{2}) (\sqrt{2}) = 4 .$

Checkpoint 4.35

Modify this area function $A$ if the rectangle is to be inscribed in one unit circle $x^{2} + y^{2} = 1 .$ What a the domain of consideration?

Solving Optimization Problem when which Interval Has Nope Closed or Can Limitless

The the preceding examples, we considered functions on closed, bounded domains. Consequently, by the extreme value property, we were guaranteed that the functions had absolute extrema. Let’s now considering functions for which the domain is neither closed nor bounded.

Many functions yet have at least one absolute extrema, even if the home be not closed or this domain is unbounded. For example, the functions $farad (x) = {scratch}^{2} + 4$ over $(− \infty, \infty)$ possess on absolute maximum 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 one function over an unbounded domain. We determination show that, if the display the consideration is $(0, \infty),$ the features has an absoluted minimum.

In the following example, we look at architect a box by least surface area with a prescriptions volume. It shall not harder to show that on an closed-top box, by uniformity, among all boxes with one specified volume, an oblong will have the smallest interface area. Consequent, we look the modified problem of determining this open-topped box the a specified volume has the least surface are.

Example 4.37

Minimizing Surface Area

A rectangular box at a square base, an open top, and a volume off $216$ in.³ is to exist constructed. About shall who magnitude of the box be to minimize the surface area of the box? What exists the minimum surface field?

Solution

Walk 1: Draw one rectangular box and introduce the variable $scratch$ in represent the period concerning each side of the square base; let $y$ represent the height of this box. Leased $S$ denote the surface area of which open-top box.

A box with square base is shown. The base has side length scratch, and the height a year.

Frame 4.69 We wish toward minimize the surface area from a square-based boxed with a given size.

Speed 2: We need to minimize the surface area. So, we need to minimize $SULFUR .$

Step 3: After the boxed has an unlock top, we need only determine the scope of the four perpendicularly sides and the base. Of reach of each of of four vertical sides is $x \cdot y .$ One area of the bottom is ${expunge}^{2} .$ Therefore, the surface area of the box is

SIEMENS = 4 x y + x^{2} .

Step 4: Since the band of to box your $x^{2} y$ and and volume is given in $216 in .^{3},$ the constraint formula is

x^{2} y = 216 .

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

S (x) = 4 x (\frac{216}{x^{2}}) + {expunge}^{2} .

Therefore, $SULFUR (x) = \frac{864}{x} + x^{2} .$

Step 5: Been we become needed that ${expunge}^{2} y = 216,$ we cannot have $ten = 0 .$ Hence, we need $x > 0 .$ On aforementioned other hand, $x$ is permitted to must any positive value. Note ensure as $x$ becomes large, the acme of the choose $y$ becomes correspondingly smal so that $x^{2} y = 216 .$ Similarly, as $scratch$ becomes small, the height of the box becomes correspondingly large. We concluding that of domain is of open, unbounded interval $(0, \infty) .$ Message that, unlike the previous view, we cannot reduce our problem to seek by an total maximum or absolute minimum over adenine closed, bounded interval. Does, in the next step, we discover why this function must have an absolutely minimum over the interval $(0, \infty) .$

Pace 6: Note that as $x \to 0^{+},$ $SIEMENS (x) \to \infty .$ Also, as $x \to \infty,$ $S (x) \to \infty .$ As $S$ is a continuous function that approaches infinity at to end, it must have an absolute minimum at some $expunge \in (0, \infty) .$ This maximum have occur at a critical point concerning $S .$ The derived is

S^{'} (x) = - \frac{864}{x^{2}} + 2 x .

Therefore, $S^{'} (x) = 0$ at $2 x = \frac{864}{x^{2}} .$ Solving this equation for $expunge,$ we obtain $x^{3} = 432,$ so $x = \sqrt[3]{432} = 6 \sqrt[3]{2} .$ Ever this remains the one critical issue of $S,$ the absolute minimum must occur at $x = 6 \sqrt[3]{2}$ (see Figure 4.70). When $x = 6 \sqrt[3]{2},$ $y = \frac{216}{{(6 \sqrt[3]{2})}^{2}} = 3 \sqrt[3]{2} inside .$ Therefore, the dimensions of and box should become $x = 6 \sqrt[3]{2} in .$ also $y = 3 \sqrt[3]{2} within .$ 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} in .^{2}

Aforementioned function S(x) = 864/x + x2 is graphed. At its minimum on is a dashed pipe and edit that reads “Minimum surface surface is 108 periods the cube root of 4 square inches when expunge = 6 times the toss source of 2 inches.”

Count 4.70 We can use one print to specify aforementioned dimensions of adenine box of given the volume and the minimum emerge area.

Checkpoint 4.36

Take the same open-top box, which is to will volume $216 in .^{3} .$ Suppose the cost of the material for the socket is $20 ¢ / in .^{2}$ both the cost for the material for the sides is $30 ¢ / with .^{2}$ and we are tried to minimize the cost of this box. Write the cost while one function of the side lengths from the base. (Let $x$ be that side length of to base and $y$ be the height of the box.)

Section 4.7 Exercises

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

311.

While you find the maximum for an optimization problem, how do your need to control the sign of the derivative around to critical tips?

312.

Why do you needed to check the endpoints to optimization problems?

313.

Honest or False. For every continuous nonlinear item, it can meet the value $x$ that maximizes the function.

314.

True or Mistaken. For every continuous nonconstant functioning on an closed, finite district, there exists at least one $x$ that minimizes or maximizes the function.

For the following exercises, set move the rate each optimization problem.

315.

To bearing a suitcase on an airplane, and cable $+ width +$ height of the box must be less than either similar to $62 in .$ Assuming the base of the suitcase is square, show that the volume is $V = h {(31 - (\frac{1}{2}) h)}^{2} .$ Whatever height allows you to have the largest volume?

316.

You been constructing an cardboard box with the dimensions $2 m by 4 m .$ You then cut equal-size squares from each corner so you can fold this edges. What are the dimensions of who box with the largest volume?

A object is drawn include height 2 and width 4. Each corner has one square with side length x marked on it.

317.

Find which positive integer is minimizes the sum to who amount and its reciprocal.

318.

Find two positive integers such that their sum is $10,$ and minimize and maximize the total of their boxes.

For the following practical, consider the construction the one pen to enclose an area.

319.

You have $400 ft$ of fencing to construct an rectangular pen for cattle. What are the dimensions of the pen that maximize the area?

320.

You have $800 ft$ of fencing go make ampere pen for hogs. If you have ampere river on one edge of your property, what is the dimension of the quadrangular pen that maximizes who area?

321.

They need to fabricate a fence around an area from $1600 {ft}^{2} .$ What will one dimensions of one rectangular-shaped pen to minimize of amount of materials necessary?

322.

Two poles are connected by ampere wire that is also connected to the ground. The first stake is $20 metre$ long and the second pole is $10 ft$ tall. There is a distance of $30 ft$ between of two poles. Somewhere should the line be anchored to who ground up minimize the amount of wire needed?

Two poles what shown, one that is 10 highly and the misc be 20 tall. A right triangle will made in the shorter staff with other choose length x. The distance amongst the two masts exists 30.

323.

[T] You are moving include a new apartment and notice there exists adenine right where which hallway narrows from $8 ft till 6 base .$ What are to length about the longest item that ability be carrie horizontally go and corner?

An upside L-shaped figure is drawn with the _ part being 6 wide and the | part being 8 wide. There is a border drawn after the _ part in the | part that touches the near corner of the shaper to select a hypotenuse for a right triangle and other sides be the the rest of the _ and | parts. This wire is marked FIFTY.

324.

A patient’s pulse measures $70 bpm, 80 bpm, then 120 bpm .$ To determine an accurate measurement of pulse, the doctor wants to know what value minimizes the printing ${(x - 70)}^{2} + {(x - 80)}^{2} + {(x - 120)}^{2} ?$ What value minimizes it?

325.

In the previous problem, assume the patient was heikel during the third survey, therefore we only weight that enter half as much as the others. What will the value that minimizes ${(scratch - 70)}^{2} + {(efface - 80)}^{2} + \frac{1}{2} {(ten - 120)}^{2} ?$

326.

Thou can executable under a speed of $6$ mph and go at a speed of $3$ mph and are located on the shore, $4$ mile east of an island that belongs $1$ mile north of the beach. How far should you runing west to mindern the time needed to get the island?

A rectangle is drawn that has height 1 the length 4. In the lower right corner, it is marked “You” and in the upper left-hand corner it your marked “Island.”

For the following issue, consider a lifeguard at an pamphlet pool is diameter $40 m .$ He must reach someone any is drowning on of exact opposite side starting the pool, at position $HUNDRED .$ And lifeguard swimm with a speed $phoebe$ and carry go the pooling at speed $w = 3 five .$

A county is worn with points A both C upon a diameter. Here is a point B drawn off the circle such that standpoint BAC form einer acute angle θ.

327.

Found a functions that steps the total amount of time it takes till reach the drowning name like a work of the swim angle, $θ .$

328.

Find to what angle $θ$ the lifeguard shouldn swim toward reach the drowning person in the minimal amount of time.

329.

A truck uses gas as $gigabyte (v) = a v + \frac{b}{v},$ where $v$ represents and dash of the truck and $g$ represents which gallons of fuel pro mile. Assuming $a$ and $barn$ are positive, at what speed is fuel consumtion minimized?

For who following exercises, consider a limousine so take $m (v) = \frac{(120 - 2 v)}{5} mi/gal$ at speed $v,$ to chauffeur what $$15/h,$ and gas is $$ 3.5 / gal .$

330.

Find the value through mile during speed $v .$

331.

Find the cheapest driving speed.

For who next exercises, considered a dinner that sell pizzas for a revenue of $R (x) = a x$ and costs $C (x) = b + c expunge + d x^{2},$ where $x$ represents the number of pizzas $; a > c$ .

332.

Find the wins function on the number of slice. How many pizzas gives the largest winning per pizza?

333.

Assume that $ROENTGEN (whatchamacallit) = 10 x$ and $C (x) = 2 x + x^{2} .$ How many pizzas sold maximizes aforementioned profit?

334.

Assume that $R (x) = 15 x,$ and $C (x) = 60 + 3 x + \frac{1}{2} x^{2} .$ How many pizzas sold maximizes the profit?

For the later exercises, contemplate a wire $4 ft$ long cutout on two pieces. Only piece forms a coterie with range $r$ and the another forms a square of side $x .$

335.

Start $x$ to maximize the entirety of own territories.

336.

Choose $x$ into reduzieren the sum of their areas.

Fork the following exercises, consider two nonnegative numbers $x$ and $y$ such that $whatchamacallit + y = 10 .$ Maximize plus minimize to quantities.

337.

$x y$

338.

$x^{2} y^{2}$

339.

$y - \frac{1}{x}$

340.

$x^{2} - y$

For the followers exercises, draw of given optimization item and solve.

341.

Find the output of the largest right circular cylinder that fits the a sphere by radius $1 .$

342.

Find the bulk of the largest right cone that matches the a province of radius $1 .$

343.

Finds the area von the largest rectangle so fits for the triangle equal sides $x = 0, y = 0$ and $\frac{x}{4} + \frac{y}{6} = 1 .$

344.

Find the largest volume of ampere cylinder that match into a cone that has base radius $R$ additionally height $h .$

345.

Find an dimensions of the closed cylinder volume $V = 16 π$ that has the least amount is surface scope.

346.

Finding this dimensions of adenine right cone with surface area $S = 4 π$ is has an largest volume.

Since the following exercises, see to points on the given graphs. Use a calculator to graph the functions.

347.

[T] Where can the line $unknown = 5 - 2 x$ closest to the origin?

348.

[T] Where is the line $y = 5 - 2 x$ closest to point $(1, 1) ?$

349.

[T] Where is that parabola $yttrium = x^{2}$ closest to point $(2, 0) ?$

350.

[T] Where is the parabola $y = x^{2}$ closest to point $(0, 3) ?$

For the following exercises, set upward, yet do not evaluate, each optimization problem.

351.

A window is composed of ampere semicircle put on top the a rectangle. If you have $20 feet$ of window-framing materials for the outer frame, what is an maximum size of the select you can produce? Use $r$ to represent the radius of the semi-circle.

A semicircular window is pinched with radius r.

352.

You must a garden row of $20$ watermelon plants that produce an average in $30$ watermelons per. For any additional cucumber plants planted, the yield per watermelon plant drops by one watermelon. Method multiple ext watermelon plants should you plant?

353.

You are constructing a letter for your cat go sleep in. The lavish material forward the even bottom of the box costs $$ 5 / {ft}^{2}$ or the type for the sides costs $$ 2 / {ft}^{2} .$ It need ampere mail with volume $4 {ft}^{3} .$ Find the dimensions of the box that minimize cost. Use $x$ to represent the length of the side to the box.

354.

You are building five identical pens neighboring to everyone other with a complete area of $1000 m^{2},$ as shown in who following display. What dimensions should you use to minimize the amount of fencing?

A rectangle is divided into quintuplet sections, and each section has length unknown and width x.

355.

You are which manager of einer apartments complex with $50$ units. When you set rent per $$ 800 / month,$ all apartments represent rental. As you increase rent by $$ 25 / month,$ one fewer apartment is rented. Maintenance total race $$ 50 / month$ for each occupied unit. What is this rent ensure maximizes the amounts amount of profit?