3.6 applications in optimization

Applications in Optimization

In this section, we solve various extrema problems in real world applications. These problems are referred to as “optimization problems” in mathematics, and derivatives play a major role in such problems.


extrema-existence–problem.

In this section, we solve various extrema problems in real world applications. These problems are referred to as “optimization problems” in mathematics, and derivatives play a major role in such problems. We begin with an


extrema-existence–problem. Let y be a function defined over an interval V, y may or may not have any extremum in V.



extrema-existence–problem. Let y be a function defined over an interval V, y may or may not have any extremum in V. Here are two examples of well behaved functions that don’t have extrema.



extrema-existence–problem. Let y be a function defined over an interval V, y may or may not have any extremum in V. Here are two examples of well behaved functions that don’t have extrema. Let y = tan(x) over the interval V = (–π/2, π/2) as shown.



y=tan(x)

–π/2 π/2

x

y

extrema-existence–problem. Let y be a function defined over an interval V, y may or may not have any extremum in V. Here are two examples of well behaved functions that don’t have extrema. Let y = tan(x) over the interval V = (–π/2, π/2) as shown. There is no extremum in V because V is open.



y=tan(x)

–π/2 π/2

x

y

Even if we close the interval V, it’s still not enough to guarantee the existence of an extremum.




x

y=g(x)

For example, let y = g(x) be the function over the closed interval U = [–π/2, π/2] as shown.

–π/2 π/2

y (0, 1)

(0, –1)



x

y=g(x)

For example, let y = g(x) be the function over the closed interval U = [–π/2, π/2] as shown. Note that even though g(x) is bounded between

–π/2 π/2

y (0, 1)

(0, –1)

1 and –1, it still doesn’t have any extremum in U.



x

y=g(x)


–π/2 π/2

y (0, 1)

(0, –1)

1 and –1, it still doesn’t have any extremum in U.(Extrema Theorem for Continuous Functions)



x

y=g(x)


–π/2 π/2

y (0, 1)

(0, –1)

1 and –1, it still doesn’t have any extremum in U.(Extrema Theorem for Continuous Functions)Let y = f(x) be a continuous function defined over a closed interval V = [a, b], then both the absolute max. and the absolute min. exist in V.



x

y=g(x)


–π/2 π/2

y (0, 1)

(0, –1)

1 and –1, it still doesn’t have any extremum in U.(Extrema Theorem for Continuous Functions)Let y = f(x) be a continuous function defined over a closed interval V = [a, b], then both the absolute max. and the absolute min. exist in V. Furthermore, the absolute extrema must occur where f '(x) = 0,



x

y=g(x)


–π/2 π/2

y (0, 1)

(0, –1)

1 and –1, it still doesn’t have any extremum in U.(Extrema Theorem for Continuous Functions)Let y = f(x) be a continuous function defined over a closed interval V = [a, b], then both the absolute max. and the absolute min. exist in V. Furthermore, the absolute extrema must occur where f '(x) = 0, or where f'(x) is UDF, or they occur at the end points {a, b}.

Applications in OptimizationHere are examples of each type of extrema.


A.

a xb c


A.

a xb

In A. The absolute maximum occurs at x = c where f '(c) = 0, and the absolute minimum occurs at the end point x = b.

c

A f'=0 ab. max.

An end–point ab. min.


A.

a xb

B.

a xb


c

c d

A f'=0 ab. max.



A.

a xb

B.

a xb


c

c d

In B. The absolute maximum occurs at x = c where f '(c) is UDF, and the absolute minimum occurs at x = dwhere f '(d) = 0.

A f'=0 ab. max.


A UDF– f' ab. max.

A f'=0 ab. min.


A.

a xb

B.

a xb


c

c d

In B. The absolute maximum occurs at x = c where f '(c) is UDF, and the absolute minimum occurs at x = dwhere f '(d) = 0.

A f'=0 ab. max.


A UDF– f' ab. max.

A f'=0 ab. min.

Hence we have to consider points of all three cases when solving for extrema.

Applications in OptimizationExample A. Let y be defined as the following.

x2 for 1 > xy =

2 – x for 1 ≤ x


x2 for 1 > xy =

2 – x for 1 ≤ x(1, 1)

x

y


x2 for 1 > xy =

2 – x for 1 ≤ x(1, 1)

x

y

Find the extrema of y over the interval [–1 , ½ ].


x2 for 1 > xy =

2 – x for 1 ≤ x(1, 1)

x

y

Find the extrema of y over the interval [–1 , ½ ].The end–point values are f(–1) = 1 and f(½) = ¼.


x2 for 1 > xy =

2 – x for 1 ≤ x(1, 1)

x

y


x

y

The end–point values are f(–1) = 1 and f(½) = ¼.

–1 ½

(–1, 1)

(½, ¼ )


x2 for 1 > xy =

2 – x for 1 ≤ x(1, 1)

x

y


x

y

The end–point values are f(–1) = 1 and f(½) = ¼. y' = 0 at x = 0 so that (0, 0) is an f'=0–type minimum.

–1 ½

(–1, 1)

(½, ¼ )


x2 for 1 > xy =

2 – x for 1 ≤ x(1, 1)

x

y


x

y

Within the interval [–1 , ½ ], there is no point of the f'–UDF–type. Hence the ab. max. is at (–1, 1) and the ab. min is at (0, 0).

The end–point values are f(–1) = 1 and f(½) = ¼. y' = 0 at x = 0 so that (0, 0) is an f'=0–type minimum.

–1 ½

(–1, 1)

(½, ¼ )

Applications in OptimizationThe key here is that if we are to find extrema of a given continuous function over a closed interval we have to obtain all three types of points described above, then select the correct solutions.


x

yFor instance, within [–½, 3], x = 0 is an f'=0–type solution.The point (1, 1) at x = 1 is an f'–UDF–type solution.

3–½

(1, 1)

The key here is that if we are to find extrema of a given continuous function over a closed interval we have to obtain all three types of points described above, then select the correct solutions.


x

yFor instance, within [–½, 3], x = 0 is an f'=0–type solution.The point (1, 1) at x = 1 is an f'–UDF–type solution. The end–point values are f(–½) = ¼ and f(3) = –1.

3–½

(1, 1)


(3, –1)


x

yFor instance, within [–½, 3], x = 0 is an f'=0–type solution.The point (1, 1) at x = 1 is an f'–UDF–type solution. The end–point values are f(–½) = ¼ and f(3) = –1.By inspection, the ab. max. is at (1, 1) and the ab. min is at (3, –1).

3–½

(1, 1)


(3, –1)


x

yFor instance, within [–½, 3], x = 0 is an f'=0–type solution.The point (1, 1) at x = 1 is an f'–UDF–type solution. The end–point values are f(–½) = ¼ and f(3) = –1.By inspection, the ab. max. is at (1, 1) and the ab. min is at (3, –1).

3–½

(1, 1)


(3, –1)

In the applied problems below, make sure all such relevant points are considered.

Applications in OptimizationDistance Problems


The distance formula is a square–root formula whose radicand is always nonnegative. To find the extrema of a distance relation in the form of “√u(x)”, we drop the square–root and look for the extrema of u(x) instead.

Distance Problems



Distance Problems

Example B. Find the coordinate of the point on the line y = 2x that is closest to the point (1, 0).



Distance Problems


x

y=2x

(1, 0)

y

D



Distance Problems


x

y=2x

(1, 0)

y

(x, 2x)

A generic point on the line is (x, 2x) and its distance to the point (1, 0) is D = [(x – 1)2 + (2x – 0 )2]½.

D



Distance Problems


x

y=2x

(1, 0)

y

(x, 2x)

A generic point on the line is (x, 2x) and its distance to the point (1, 0) is D = [(x – 1)2 + (2x – 0 )2]½. We want to find the x value that gives the minimal D. Note that D is defined for all numbers.

D


x

(1, 0)

(x, 2x)

We minimize the radicandu = D2 = (x – 1)2 + (2x – 0 )2 = 5x2 – 2x + 1 instead.

y

D


x

(1, 0)

(x, 2x)

We minimize the radicandu = D2 = (x – 1)2 + (2x – 0 )2 = 5x2 – 2x + 1 instead. Set u' = 10x – 2 = 0. We get x = 1/5, and the point is (1/5, 2/5).

y

D


x

(1, 0)

(x, 2x)

We minimize the radicandu = D2 = (x – 1)2 + (2x – 0 )2 = 5x2 – 2x + 1 instead. Set u' = 10x – 2 = 0. We get x = 1/5, and the point is (1/5, 2/5).From the geometry it’s obvious (1/5, 2/5) is the closest point to (1, 0).

y

D


x

(1, 0)

(x, 2x)


½

y

D

Similarly if we wish to find the extrema of eu(x) or In(u(x)), we may find the extrema of u(x) instead. This is true because like sqrt(x), eu(x) and In(u(x)) are increasing functions, i.e. if s < t, then es < et,so the extrema of eu(x) are the same as u(x).


x

(1, 0)

(x, 2x)


½

y

D

Similarly if we wish to find the extrema of eu(x) or In(u(x)), we may find the extrema of u(x) instead. This is true because like sqrt(x), eu(x) and In(u(x)) are increasing functions, i.e. if s < t, then es < et,so the extrema of eu(x) are the same as u(x). In fact, if f(x) is an increasing function, then the extrema of f(u(x)) are the same as the ones of u(x).

Applications in Optimization(What is the corresponding statement about the extrema of f(u(x)) and u(x) if f(x) is a decreasing function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)

Applications in Optimization(What is the corresponding statement about the extrema of f(u(x)) and u(x) if f(x) is a decreasing function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)Area ProblemsIn area optimization problems, the main task is to express the area in question in a single well chosen variable x as A(x).

Applications in Optimization(What is the corresponding statement about the extrema of f(u(x)) and u(x) if f(x) is a decreasing function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)Area ProblemsIn area optimization problems, the main task is to express the area in question in a single well chosen variable x as A(x). If x represents a specific measurement, as opposed to a coordinate, then x must be nonnegative (often it’s also bounded above).

Applications in Optimization(What is the corresponding statement about the extrema of f(u(x)) and u(x) if f(x) is a decreasing function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)Area ProblemsIn area optimization problems, the main task is to express the area in question in a single well chosen variable x as A(x). If x represents a specific measurement, as opposed to a coordinate, then x must be nonnegative (often it’s also bounded above). In such cases consider the significance when x = 0 (or it’s other boundary value). Often one of the extrema is at the boundary, and we are looking for the other extremum. A drawing is indispensable in any geometric problem.

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible? x

x/2 y

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible?The enclosed area is A = xy.

x

x/2 y

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible?The enclosed area is A = xy. The total linear lengthis 120, that is, 3y + 2½ x = 120

x

x/2 y

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible?The enclosed area is A = xy. The total linear lengthis 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6.

x

x/2 y

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible?The enclosed area is A = xy. The total linear lengthis 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6. So A(x) = x(40 – 5x/6) = 40x– 5x2/6.

x

x/2 y

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible?The enclosed area is A = xy. The total linear lengthis 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6. So A(x) = x(40 – 5x/6) = 40x– 5x2/6. At the boundary values x = 0, 48 (why?) we have the absolute minimal A = 0.

x

x/2 y

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible?The enclosed area is A = xy. The total linear lengthis 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6. So A(x) = x(40 – 5x/6) = 40x– 5x2/6. At the boundary values x = 0, 48 (why?) we have the absolute minimal A = 0. Setting A'(x) = 40 – 5x/3 = 0, we get x = 24

x

x/2 y

Applications in OptimizationExample C. Farmer Fred wants to raise chickens, ducks and pigs. He has 120 yards of fence to build rectangular regions as shown to separate the animals. What is the maximal enclosed area that is possible?The enclosed area is A = xy. The total linear lengthis 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6. So A(x) = x(40 – 5x/6) = 40x– 5x2/6. At the boundary values x = 0, 48 (why?) we have the absolute minimal A = 0. Setting A'(x) = 40 – 5x/3 = 0, we get x = 24 which yields y = 20. So 20×24 = 480 yd2 must be the maximum area.

x

x/2 y

Applications in OptimizationIn fact for any specified partition of the rectangle, the maximal area always occurs when the total–vertical–fence–length is equal to the total–horizontal–fence–length. That is, using half of the fence for vertical dividers and the other half for the horizontal dividers will give the maximum enclosed area.Hence if farmer Joe is to build an enclosure with 120 yd of fence as shown, then the answer is x = 60/3½ = 120/7, y = 60/3 = 20 with the maximum possible area of 120/7×20 ≈ 343 yd2. x

x/2

y

Applications in OptimizationExample D. Farmer Joe wishes to build an open cylindrical water tank made from sheet–metal that will hold 120π yd3 of water. What is the least amount of sheet–metal (in yd2) needed?

120 yd3

h

r


120 yd3

h

r

The volume of a cylinder is V = Bh where B = area of the base and h = height


120 yd3

h

r

The volume of a cylinder is V = Bh where B = area of the base and h = height so we have120π = πr2hor h = 120/r2.


120 yd3

h

r

The volume of a cylinder is V = Bh where B = area of the base and h = height so we have120π = πr2hor h = 120/r2. The surface area of the cylinder consists of a circularbase and the circular wall.


120 yd3

h

r

The volume of a cylinder is V = Bh where B = area of the base and h = height so we have120π = πr2hor h = 120/r2. The surface area of the cylinder consists of a circularbase and the circular wall. The base area is πr2.


120 yd3

h

r

The volume of a cylinder is V = Bh where B = area of the base and h = height so we have120π = πr2hor h = 120/r2. The surface area of the cylinder consists of a circularbase and the circular wall. The base area is πr2. Unroll and flatten the wall, it is an h x 2πr rectangular sheet.


120 yd3

h

r

The volume of a cylinder is V = Bh where B = area of the base and h = height so we have120π = πr2hor h = 120/r2. The surface area of the cylinder consists of a circularbase and the circular wall. The base area is πr2. Unroll and flatten the wall, it is an h x 2πr rectangular sheet. Hence the total surface is S = πr2+ 2πrh,or S = πr2 + 2πr( )120

r2


120 yd3

h

r

The volume of a cylinder is V = Bh where B = area of the base and h = height so we have120π = πr2hor h = 120/r2. The surface area of the cylinder consists of a circularbase and the circular wall. The base area is πr2. Unroll and flatten the wall, it is an h x 2πr rectangular sheet. Hence the total surface is S = πr2+ 2πrh,or S = πr2 + 2πr( ) = πr2 + 240π/r = π(r2 + 240/r)120

r2


120 yd3

h

r

Setting S' = 2π(r – 120/r2) = 0,dividing both sides by 2π and clearing the denominator yields r3 – 120 = 0


120 yd3

h

r


or that r = 1201/3, which gives the absolute minimal (why?) surface


120 yd3

h

r



of π(1202/3 + ) = π 360/1201/3 2401201/3 yd2.


120 yd3

h

r



of π(1202/3 + ) = π 360/1201/3 2401201/3 yd2.

If the geometric object has a variable angle θ, then θ may be utilized as the independent variable.


120 yd3

h

r



of π(1202/3 + ) = π 360/1201/3 2401201/3 yd2.

If the geometric object has a variable angle θ, then θ may be utilized as the independent variable.Example E. Find the isosceles triangle having two sides of length 4 with the largest area as shown.

44

Applications in OptimizationThere are two observations that will make the algebra cleaner. 4

4

Applications in OptimizationThere are two observations that will make the algebra cleaner. First let’s take advantage of the symmetry. Draw a perpendicular line as shown, it cuts the triangle

44

into two mirror image right triangles.


44

into two mirror image right triangles. We may maximize one of the right triangles instead, i.e. to find the largest area possible of a right triangle with hypotenuse equal to 4.

4

a

bA


44

into two mirror image right triangles. We may maximize one of the right triangles instead, i.e. to find the largest area possible of a right triangle with hypotenuse equal to 4.

4

a

b

Second, instead of expressing the area A in terms of the measurements of the legs a and b, we express A in terms of the angle θ as shown.

A

θ

Applications in OptimizationA = ½ ab so in terms of θ we have

4

a

bA

θ

A = ½ 4sin(θ) 4cos(θ) = 8 sin(θ) cos(θ)


Note that 0 ≤ θ ≤ π/2.

4

a

bA

θ



Note that 0 ≤ θ ≤ π/2.Set A'(θ) = 8(cos2(θ) – sin2 (θ)) = 0

4

a

bA

θ




4

a

bA

θ


However, the only solution with 0 ≤ θ ≤ π/2 is θ = π/4Hence cos(θ) = ±sin (θ)



4

a

b

which must be a maximum (why?). So the largest possible isosceles triangle in question is the right isosceles triangle as shown with area 2A = ½ (4 ×4) = 8.

A

θ


However, the only solution with 0 ≤ θ ≤ π/2 is θ = π/4Hence cos(θ) = ±sin (θ)

44 π/2

Applications in OptimizationExample F. A duck can walk 3 ft/sec and swim2 ft/sec at its top speed. It sees a piece of bread across the pool as shown. What is the path it should take to get to the bread as soon as possible?

40 ft

30 ft

pool


40 ft

30 ft

poolSelect the x as shown in the picture.

x


40 ft

30 ft

poolSelect the x as shown in the picture. (The selection of the correct x is important. The wrong choice leads to tangled algebra as is often the case in these problems.)

x


40 ft

30 ft

poolSelect the x as shown in the picture. (The selection of the correct x is important. The wrong choice leads to tangled algebra as is often the case in these problems.)

Hence if your choice of x leads to nowhere, try another choice.

x


40 ft

30 ft

pool

x

Note that 0 ≤ x ≤ 40, where x = 0 means the duck walks aroundthe pool and x = 40 means the duck swims all the way.

Select the x as shown in the picture. (The selection of the correct x is important. The wrong choice leads to tangled algebra as is often the case in these problems.)


40 ft

30 ft

pool

x(x2 + 900)½

For x ≠ 0, the duck swims a distance of (x2 + 900)½


40 ft

30 ft

pool

x(x2 + 900)½

40 – x

For x ≠ 0, the duck swims a distance of (x2 + 900)½ and walks a distance of (40 – x),


40 ft

30 ft

pool

x

and the total time t it takes is(x2 + 900)½

40 – xt(x) = (x2 + 900)½

2 +(40 – x)

3


swimming time walking time


40 ft

30 ft

pool

x


40 – xt(x) = (x2 + 900)½

2 +(40 – x)

3Set t'(x) = 2(x2 + 900)½

x – 31


= 0


40 ft

30 ft

pool

x


40 – xt(x) = (x2 + 900)½

2 +(40 – x)

3Set t'(x) = 2(x2 + 900)½

x – 31


= 0

We have x = 31

2(x2 + 900)½


40 ft

30 ft

pool

x


40 – xt(x) = (x2 + 900)½

2 +(40 – x)

3Set t'(x) = 2(x2 + 900)½

x – 31


= 0

We have x = 31

2(x2 + 900)½ 3x = 2(x2 + 900)½


40 ft

30 ft

pool

x


40 – xt(x) = (x2 + 900)½

2 +(40 – x)

3Set t'(x) = 2(x2 + 900)½

x – 31


= 0

We have x = 31

2(x2 + 900)½ 3x = 2(x2 + 900)½ 9x2 = 4(x2 + 900)


40 ft

30 ft

pool

x


40 – xt(x) = (x2 + 900)½

2 +(40 – x)

3Set t'(x) = 2(x2 + 900)½

x – 31


= 0

We have x = 31

2(x2 + 900)½ 3x = 2(x2 + 900)½ 9x2 = 4(x2 + 900)

5x2 = 3600 x = ±12√5Discarding the negative answer, x =12√5 ≈ 26.8 ft.

Applications in OptimizationWhen x = 0, the time it takes is not t(0) because t(x) assume the duck swims at least 30 ft,

Applications in OptimizationWhen x = 0, the time it takes is not t(0) because t(x) assume the duck swims at least 30 ft, as it would be the case if the duck is across a river from the bread instead of a pool.

Applications in OptimizationWhen x = 0, the time it takes is not t(0) because t(x) assume the duck swims at least 30 ft, as it would be the case if the duck is across a river from the bread instead of a pool. Let’s examine the three points.


When x = 40 the duck swims all the way and it takes t(40) = 50/2 = 25 seconds.

When x = 0, the time it takes is not t(0) because t(x) assume the duck swims at least 30 ft, as it would be the case if the duck is across a river from the bread instead of a pool. Let’s examine the three points. When x = 0, the duck walks 70 ft and it takes 70/3 = 23 1/3 seconds.

At the critical point, t(12√5) ≈ 24.5 seconds.




At the critical point, t(12√5) ≈ 24.5 seconds.Therefore the duck should walk around the pool.




At the critical point, t(12√5) ≈ 24.5 seconds.Therefore the duck should walk around the pool.Remarks1. As noted that if the duck is across a river instead of a pool, then indeed x = 12√5 is the absolute minimum because t(0) = swim + walk = 28 1/3 sec.

Applications in Optimization2. Just because the duck walks faster does not automatically means it should always walk around the pool. We may think about this by imagining the duck is slightly faster in walking, say at 2.01 ft/sec,than swimming at 2 ft/sec. Then of course the duck should swim toward the bread instead of wasting time walking around. In fact the closer the walking speed is to the swimming speed, the more the duck should swim toward the bread.

Back to math–265 pg

http://www.lahc.edu/math/cal2/math_265.html



3.6 applications in optimization

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