Maximizing The Volume Of A Box

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four. Applications of Derivatives

four.7 Applied Optimization Issues

Learning Objectives

  • Ready and solve optimization problems in several applied fields.

1 common awarding of calculus is calculating the minimum or maximum value of a office. For example, companies often want to minimize production costs or maximize revenue. In manufacturing, information technology is often desirable to minimize the corporeality of textile used to package a product with a certain volume. In this section, nosotros show how to fix upward these types of minimization and maximization problems and solve them by using the tools developed in this chapter.

Solving Optimization Bug over a Closed, Bounded Interval

The basic idea of the
optimization problems

that follow is the aforementioned. Nosotros have a particular quantity that nosotros are interested in maximizing or minimizing. However, nosotros also accept some auxiliary condition that needs to be satisfied. For case, in (Effigy), nosotros are interested in maximizing the area of a rectangular garden. Certainly, if we proceed making the side lengths of the garden larger, the area volition continue to get larger. However, what if we take some restriction on how much fencing we tin can use for the perimeter? In this example, we cannot brand the garden as big as we like. Let’southward await at how we tin can maximize the area of a rectangle subject to some constraint on the perimeter.

Maximizing the Area of a Garden

A rectangular garden is to exist synthetic using a rock wall as one side of the garden and wire fencing for the other iii sides ((Figure)). Given 100 ft of wire fencing, make up one’s mind the dimensions that would create a garden of maximum area. What is the maximum surface area?

Solution

Let

denote the length of the side of the garden perpendicular to the stone wall and
y
denote the length of the side parallel to the rock wall. Then the expanse of the garden is

A=x·y.

We want to find the maximum possible surface area discipline to the constraint that the total fencing is
100\text{ft}.
From (Figure), the total amount of fencing used volition exist
2x+y.
Therefore, the constraint equation is

2x+y=100.

Solving this equation for
y,
we have
y=100-2x.
Thus, nosotros tin can write the expanse equally

A(x)=x·(100-2x)=100x-2{x}^{2}.

Earlier trying to maximize the surface area part
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
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                  and<br />
                  <img src=
if
y>0,” title=”Rendered by QuickLaTeX.com” height=”16″ width=”46″><br />
                  then<br />
                  <img loading=
Therefore, we are trying to make up one’s mind the maximum value of
A(x)
for

over the open interval
(0,50).
We do not know that a function necessarily has a maximum value over an open interval. Withal, nosotros 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 airtight interval
\left[0,50\right].
If the maximum value occurs at an interior indicate, so nosotros have found the value

in the open up interval
(0,50)
that maximizes the area of the garden. Therefore, we consider the following trouble:

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

Every bit mentioned earlier, since
A
is a continuous function on a closed, divisional interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or disquisitional points. At the endpoints,
A(x)=0.
Since the area is positive for all

in the open interval
(0,50),
the maximum must occur at a critical signal. Differentiating the function
A(x),
we obtain

{A}^{\prime }(x)=100-4x.

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

Determine the maximum expanse if nosotros want to make the same rectangular garden as in (Effigy), but nosotros have 200 ft of fencing.

Solution

The maximum expanse is
5000{\text{ft}}^{2}.

Now let’s await at a general strategy for solving optimization problems similar to (Figure).

Trouble-Solving Strategy: Solving Optimization Problems

  1. Introduce all variables. If applicable, draw a figure and label all variables.
  2. Decide 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).
  3. Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more one variable.
  4. Write whatsoever equations relating the independent variables in the formula from step 3. Use these equations to write the quantity to be maximized or minimized every bit a function of ane variable.
  5. Identify the domain of consideration for the office in stride 4 based on the concrete trouble to be solved.
  6. Locate the maximum or minimum value of the role from footstep 4. This stride typically involves looking for critical points and evaluating a role at endpoints.

Now let’s utilize this strategy to maximize the volume of an open-pinnacle box given a constraint on the amount of textile to be used.

Maximizing the Volume of a Box

An open-tiptop box is to exist made from a 24 in. past 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 exist cut out of each corner to get a box with the maximum volume?

Solution

Pace ane: Let

exist the side length of the square to be removed from each corner ((Figure)). Then, the remaining four flaps can be folded up to class an open-top box. Permit
V
be the volume of the resulting box.

Step ii: We are trying to maximize the volume of a box. Therefore, the trouble is to maximize
V.

Step three: As mentioned in step 2, are trying to maximize the book of a box. The volume of a box is
V=L·W·H,
where
L,W,\text{ and }H
are the length, width, and height, respectively.

Pace 4: From (Effigy), we see that the height of the box is

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.

Footstep 5: To determine the domain of consideration, let’s examine (Figure). Certainly, we demand
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                  Furthermore, the side length of the square cannot exist 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, nosotros are trying to make up one’s mind whether there is a maximum volume of the box for<br />
                  <img fifu-featured=
over the open interval
(0,12).
Since
V
is a continuous function over the closed interval
\left[0,12\right],
nosotros know
V
volition accept an absolute maximum over the closed interval. Therefore, nosotros consider
V
over the airtight interval
\left[0,12\right]
and check whether the accented maximum occurs at an interior point.

Step 6: Since
V(x)
is a continuous function over the airtight, bounded interval
\left[0,12\right],
V
must have an absolute maximum (and an absolute minimum). Since
V(x)=0
at the endpoints and
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                  for<br />
                  <img loading=
the maximum must occur at a disquisitional point. The derivative is

{V}^{\prime }(x)=12{x}^{2}-240x+864.

To notice the critical points, nosotros 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{±}\sqrt{{(-20)}^{2}-4(1)(72)}}{2}=\frac{20\text{±}\sqrt{112}}{2}=\frac{20\text{±}4\sqrt{7}}{2}=10\text{±}2\sqrt{7}.

Since
10+2\sqrt{7}
is not in the domain of consideration, the only critical point we demand to consider is
10-2\sqrt{7}.
Therefore, the volume is maximized if we allow
x=10-2\sqrt{7}\text{in}.
The maximum book is
V(10-2\sqrt{7})=640+448\sqrt{7}\approx 1825\text{in}{.}^{3}
equally 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.”
Figure 4.
Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.

Watch a video about optimizing the volume of a box.

Minimizing Travel Time

An island is
2\text{mi}
due north of its closest bespeak along a straight shoreline. A visitor is staying at a cabin on the shore that is
6\text{mi}
west of that bespeak. The visitor is planning to get from the cabin to the island. Suppose the company runs at a rate of
8\text{mph}
and swims at a rate of
3\text{mph}.
How far should the visitor run earlier swimming to minimize the time it takes to attain the isle?

Solution

Step 1: Permit

be the distance running and let
y
exist the distance swimming ((Figure)). Let
T
be the time it takes to get from the cabin to the island.

Step ii: The trouble is to minimize
T.

Footstep 3: To find the time spent traveling from the motel to the island, add together the time spent running and the time spent pond. Since Altitude
=
Rate
×
Time
(D=R×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}.

Stride iv: From (Figure), 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 past the part

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

Step 5: From (Figure), we come across that
0\le x\le 6.
Therefore,
\left[0,6\right]
is the domain of consideration.

Stride half-dozen: Since
T(x)
is a continuous function over a airtight, bounded interval, information technology has a maximum and a minimum. Allow’southward begin past looking for any disquisitional points of
T
over the interval
\left[0,6\right].
The derivative is

{T}^{\prime }(x)=\frac{1}{8}-\frac{1}{2}\frac{{\left[{(6-x)}^{2}+4\right]}^{-1\text{/}2}}{3}·2(6-x)=\frac{1}{8}-\frac{(6-x)}{3\sqrt{{(6-x)}^{2}+4}}.

If
{T}^{\prime }(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, nosotros come across that if

satisfies this equation, and so

must satisfy

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

which implies

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

We conclude that if

is a critical point, then

satisfies

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

Therefore, the possibilities for critical points are

x=6\text{±}\frac{6}{\sqrt{55}}.

Since
x=6+6\text{/}\sqrt{55}
is not in the domain, it is not a possibility for a critical indicate. On the other mitt,
x=6-6\text{/}\sqrt{55}
is in the domain. Since nosotros squared both sides of (Figure) to arrive at the possible critical points, it remains to verify that
x=6-6\text{/}\sqrt{55}
satisfies (Effigy). Since
x=6-6\text{/}\sqrt{55}
does satisfy that equation, we conclude that
x=6-6\text{/}\sqrt{55}
is a critical bespeak, 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, nosotros conclude that
T
has a local minimum at
x\approx 5.19
mi.

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

Maximizing Acquirement

A automobile rental company charges its customers
p
dollars per twenty-four hour period, 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 client to maximize acquirement?

Solution

The visitor should charge
$75
per car per day.

Maximizing the Surface area of an Inscribed Rectangle

A rectangle is to be inscribed in the ellipse

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

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

Solution

Footstep 1: For a rectangle to exist inscribed in the ellipse, the sides of the rectangle must exist parallel to the axes. Allow
L
be the length of the rectangle and
W
be its width. Let
A
be the surface 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.
Figure 7.
Nosotros want to maximize the area of a rectangle inscribed in an ellipse.

Pace 2: The problem is to maximize
A.

Footstep three: The surface area of the rectangle is
A=LW.

Step four: Let
(x,y)
exist the corner of the rectangle that lies in the outset quadrant, as shown in (Effigy). Nosotros tin can write length
L=2x
and width
W=2y.
Since
\frac{{x}^{2}}{4+{y}^{2}=1}
and
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                  we take<br />
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Therefore, the area is

A=LW=(2x)(2y)=4x\sqrt{\frac{1-{x}^{2}}{4}}=2x\sqrt{4-{x}^{2}}.

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

Pace six: Equally mentioned earlier,
A(x)
is a continuous function over the closed, divisional interval
\left[0,2\right].
Therefore, it has an accented maximum (and accented minimum). At the endpoints
x=0
and
x=2,
A(x)=0.
For
0<ten<2,
A(x)>0.” title=”Rendered by QuickLaTeX.com” height=”18″ width=”74″><br />
                  Therefore, the maximum must occur at a critical signal. Taking the derivative of<br />
                  <img loading=
we obtain

\begin{array}{cc}\hfill A\prime (x)& =2\sqrt{4-{x}^{2}}+2x·\frac{1}{2\sqrt{4-{x}^{2}}}(-2x)\hfill \\ & =2\sqrt{4-{x}^{2}}-\frac{2{x}^{2}}{\sqrt{4-{x}^{2}}}\hfill \\ & =\frac{8-4{x}^{2}}{\sqrt{4-{x}^{2}}}.\hfill \end{array}

To observe disquisitional points, we need to find where
A\prime (x)=0.
We can run into that if

is a solution of

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

then

must satisfy

8-4{x}^{2}=0.

Therefore,
{x}^{2}=2.
Thus,
x=\text{±}\sqrt{2}
are the possible solutions of (Figure). Since nosotros are considering

over the interval
\left[0,2\right],
x=\sqrt{2}
is a possibility for a critical point, just
x=\text{−}\sqrt{2}
is not. Therefore, we cheque whether
\sqrt{2}
is a solution of (Figure). Since
x=\sqrt{2}
is a solution of (Figure), we conclude that
\sqrt{2}
is the only critical bespeak of
A(x)
in the interval
\left[0,2\right].
Therefore,
A(x)
must have an absolute maximum at the critical indicate
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 expanse of this rectangle is
A=LW=(2\sqrt{2})(\sqrt{2})=4.

Modify the expanse office
A
if the rectangle is to exist inscribed in the unit of measurement circle
{x}^{2}+{y}^{2}=1.
What is the domain of consideration?

Solution

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

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

In the previous examples, we considered functions on airtight, bounded domains. Consequently, past the farthermost value theorem, we were guaranteed that the functions had absolute extrema. Let’southward now consider functions for which the domain is neither airtight nor divisional.

Many functions still have at to the lowest degree i absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the role
f(x)={x}^{2}+4
over
(\text{−}\infty ,\infty )
has an absolute minimum of 4 at
x=0.
Therefore, nosotros tin still consider functions over unbounded domains or open intervals and decide whether they take whatsoever absolute extrema. In the next instance, we effort to minimize a part over an unbounded domain. We will come across that, although the domain of consideration is
(0,\infty ),
the function has an accented minimum.

In the following example, we look at amalgam a box of least surface area with a prescribed volume. It is non hard to show that for a closed-elevation box, by symmetry, amongst all boxes with a specified volume, a cube volition have the smallest surface surface area. Consequently, we consider the modified trouble of determining which open up-topped box with a specified volume has the smallest expanse.

Minimizing Expanse

A rectangular box with a square base of operations, an open superlative, 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 expanse?

Solution

Footstep 1: Draw a rectangular box and introduce the variable

to represent the length of each side of the square base; let
y
represent the elevation of the box. Allow
S
denote the surface surface area of the open up-elevation box.

A box with square base is shown. The base has side length x, and the height is y.
Figure viii.
We want to minimize the surface area of a square-based box with a given volume.

Step 2: We demand to minimize the surface area. Therefore, we need to minimize
S.

Step 3: Since the box has an open height, we need only make up one’s mind 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}.

Pace 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 part of

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,
nosotros cannot have
x=0.
Therefore, we need
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                  On the other paw,<br />
                  <img fifu-featured=
is allowed to have any positive value. Note that every bit

becomes big, the peak of the box
y
becomes correspondingly small then that
{x}^{2}y=216.
Similarly, as

becomes small-scale, the peak of the box becomes correspondingly large. We conclude that the domain is the open up, unbounded interval
(0,\infty ).
Notation that, dissimilar the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. Notwithstanding, in the next step, we discover why this function must take an absolute minimum over the interval
(0,\infty ).

Stride half dozen: Note that as
x\to {0}^{+},
S(x)\to \infty .
As well, as
x\to \infty ,
S(x)\to \infty .
Since
S
is a continuous function that approaches infinity at the ends, information technology must have an accented minimum at some
x\in (0,\infty ).
This minimum must occur at a critical indicate of
S.
The derivative is

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

Therefore,
{S}^{\prime }(x)=0
when
2x=\frac{864}{{x}^{2}}.
Solving this equation for
x,
nosotros obtain
{x}^{3}=432,
then
x=\sqrt[3]{432}=6\sqrt[3]{2}.
Since this is the simply critical point of
S,
the accented minimum must occur at
x=6\sqrt[3]{2}
(meet (Figure)). 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 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}

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.”
Effigy 9.
We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area.

Key Concepts

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

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

1.
When you lot find the maximum for an optimization trouble, why exercise you need to bank check the sign of the derivative around the critical points?

Solution

The critical points tin be the minima, maxima, or neither.

2.
Why do y’all need to check the endpoints for optimization problems?

3.

True or Faux
. For every continuous nonlinear function, yous can find the value

that maximizes the office.

Solution

False;
y=\text{−}{x}^{2}
has a minimum only

four.

True or False. For every continuous nonconstant function on a closed, finite domain, there exists at least one

that minimizes or maximizes the part.

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

five.
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 height is fixed, show that the maximum book is
V=h{(31-(\frac{1}{2})h)}^{2}.
What superlative allows yous to have the largest volume?

Solution

h=\frac{62}{3}
in.

6.
You lot are amalgam a cardboard box with the dimensions
\text{2 m by 4 m}.
You lot so cut equal-size squares from each corner so you may fold the edges. What are the dimensions of the box with the largest book?


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

7.
Find the positive integer that minimizes the sum of the number and its reciprocal.

8.
Detect 2 positive integers such that their sum is x, and minimize and maximize the sum of their squares.

For the post-obit exercises, consider the construction of a pen to enclose an expanse.

nine.
Y’all have
400\text{ft}
of fencing to construct a rectangular pen for cattle. What are the dimensions of the pen that maximize the area?

Solution

100\text{ft by}100\text{ft}

10.
You lot have
800\text{ft}
of fencing to brand a pen for hogs. If yous accept a river on one side of your property, what is the dimension of the rectangular pen that maximizes the expanse?

11.
You need to construct a contend around an area of
1600\text{ft}.
What are the dimensions of the rectangular pen to minimize the corporeality of fabric needed?

Solution

40\text{ft by}40\text{ft}

12.
Two poles are connected by a wire that is as well connected to the basis. The starting time 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 2 poles. Where should the wire be anchored to the ground to minimize the amount of wire needed?


Two poles are shown, one that is 10 tall and the other is 20 tall. A right triangle is made with the shorter pole with other side length x. The distance between the two poles is 30.

13. [T]
Yous 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 particular that can be carried horizontally around the corner?


An upside L-shaped figure is drawn with the _ part being 6 wide and the | part being 8 wide. There is a line drawn from the _ part to the | part that touches the near corner of the shape to form a hypotenuse for a right triangle the other sides being the the rest of the _ and | parts. This line is marked L.

Solution

19.73\text{ft}.

fourteen.
A patient’s pulse measures
\text{70 bpm, 80 bpm, then 120 bpm}.
To make up one’s mind 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?

xv.
In the previous problem, assume the patient was nervous during the third measurement, so nosotros only weight that value half equally much as the others. What is the value that minimizes
{(x-70)}^{2}+{(x-80)}^{2}+\frac{1}{2}{(x-120)}^{2}?

Solution

84\text{bpm}

xvi.
You tin run at a speed of 6 mph and swim at a speed of 3 mph and are located on the shore, 4 miles due east of an island that is 1 mile north of the shoreline. How far should y’all run west to minimize the fourth dimension needed to reach the island?


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

For the post-obit 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.


A circle is drawn with points A and C on a diameter. There is a point B drawn on the circle such that angle BAC form an acute angle θ.

17.
Find a part that measures the total amount of time it takes to attain the drowning person as a part of the swim angle,
\theta .

Solution

T(\theta )=\frac{40\theta }{3v}+\frac{40 \cos \theta }{v}

xviii.
Notice at what angle
\theta
the lifeguard should swim to accomplish the drowning person in the to the lowest degree amount of time.

19.
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. At what speed is fuel consumption minimized?

Solution

v=\sqrt{\frac{b}{a}}

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

20.
Find the cost per mile at speed
v.

21.
Find the cheapest driving speed.

Solution

approximately
34.02\text{mph}

For the post-obit exercises, consider a pizzeria that sell pizzas for a revenue of
R(x)=ax
and costs
C(x)=b+cx+d{x}^{2},
where

represents the number of pizzas.

22.
Notice the profit function for the number of pizzas. How many pizzas gives the largest profit per pizza?

23.
Assume that
R(x)=10x
and
C(x)=2x+{x}^{2}.
How many pizzas sold maximizes the profit?

24.
Presume that
R(x)=15x,
and
C(x)=60+3x+\frac{1}{2}{x}^{2}.
How many pizzas sold maximizes the turn a profit?

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.

25.
Choose

to maximize the sum of their areas.

26.
Choose

to minimize the sum of their areas.

For the following exercises, consider 2 nonnegative numbers

and
y
such that
x+y=10.
Maximize and minimize the quantities.

27.
xy

28.
{x}^{2}{y}^{2}

29.
y-\frac{1}{x}

Solution

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

xxx.
{x}^{2}-y

For the following exercises, depict the given optimization problem and solve.

31.
Discover the volume of the largest right circular cylinder that fits in a sphere of radius one.

Solution

\frac{4\pi }{3\sqrt{3}}

32.
Find the volume of the largest right cone that fits in a sphere of radius ane.

33.
Discover the area of the largest rectangle that fits into the triangle with sides
x=0,y=0
and
\frac{x}{4}+\frac{y}{6}=1.

34.
Notice the largest volume of a cylinder that fits into a cone that has base radius
R
and height
h.

35.
Observe the dimensions of the closed cylinder volume
V=16\pi
that has the to the lowest degree amount of surface area.

Solution

r=2,h=4

36.
Detect the dimensions of a right cone with surface area
S=4\pi
that has the largest volume.

For the following exercises, consider the points on the given graphs. Utilise a calculator to graph the functions.

37. [T]
Where is the line
y=5-2x
closest to the origin?

Solution

(2,1)

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

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

Solution

(0.8351,0.6974)

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

For the following exercises, ready, simply practice not evaluate, each optimization problem.

41.
A window is composed of a semicircle placed on elevation of a rectangle. If you have
20\text{ft}
of window-framing materials for the outer frame, what is the maximum size of the window you can create? Utilise
r
to represent the radius of the semicircle.


A semicircular window is drawn with radius r.

Solution

A=20r-2{r}^{2}-\frac{1}{2}\pi {r}^{2}

42.
You have a garden row of 20 watermelon plants that produce an average of 30 watermelons apiece. For any additional watermelon plants planted, the output per watermelon plant drops by one watermelon. How many actress watermelon plants should you plant?

44.
Y’all are building v identical pens adjacent to each other with a total area of
1000{\text{m}}^{2},
as shown in the following figure. What dimensions should you utilise to minimize the corporeality of fencing?


A rectangle is divided into five sections, and each section has length y and width x.

45.
You are the manager of an apartment complex with l units. When you gear up rent at
$800\text{/}\text{month,}
all apartments are rented. Every bit yous increase rent by
$25\text{/}\text{month,}
one fewer apartment is rented. Maintenance costs run
$50\text{/}\text{month}
for each occupied unit of measurement. What is the rent that maximizes the total amount of profit?

Solution

P(x)=(50-x)(800+25x-50)

Glossary

optimization problems
problems that are solved by finding the maximum or minimum value of a function

Source: https://opentextbc.ca/calculusv1openstax/chapter/applied-optimization-problems/