Optimizasyon Problemlerinin C˘ozum unde Takip Edilecek Ad mlarkisi.deu.edu.tr/.../optimizasyonproblemleri.pdf · Optimizasyon Problemleri 4 Q’yu Ad m3’deki di ger semboller cinsinden

Optimizasyon Problemleri


Optimizasyon Problemlerinin Çözümünde Takip Edilecek Adımlar

1 Problemi Anlayınız. Birinci adım problemi iyice anlayana kadardikkatlice okumaktır. Kendinize sorunuz. Bilinmeyen nedir? Verilennicelikler nelerdir? Verilen koşullar nelerdir?

2 Bir Şema Çiziniz. Çoğu problemde bir şema çizmek ve şemaüzerinde verilenve istenen nicelikleri belirlemek yararlı olur.

3 Gösterimleri Geliştiriniz. Maksimumu ya da minimumu bulunacakolan niceliğe bir sembol atayınız (şimdilik Q diyelim). Diğerbilinmeyen nicelikler için semboller seçiniz (a, b, c, . . . , x, y) ve busembolleri şema üzerinde gösteriniz. Verilen niceliklerin ilk harflerinisemboller olarak kullanmak yararlı olabilir. Örneğin, alanı A ile,yüksekliği (ingilizcesi height) h ile ve zamanı (ingilizcesi time) t ilegösterebilirsiniz.


4 Q’yu Adım 3’deki diğer semboller cinsinden ifade ediniz.

5 Eğer Q Adım 4’deki birden fazla değişkeni olan bir fonksiyon olarakifade edildiyse verilen bilgileri kullanarak bu değişkenler arasında(denklemler şeklinde) ilişkiler bulunuz. Sonra bu denklemlerikullanarak Q’nun ifadesindeki bir değişken dışında diğer bütündeğişkenleri yok ediniz. Böylece Q, tek değişkeni x’in fonksiyonuolarak Q = f(x) şeklinde ifade edilebilir. Bu fonksiyonun tanımkümesini yazınız.

6 Önceki bölümlerde öğrenildiği gibi f ’nin mutlak maksimum ya daminimum değerini bulunuz. Özel olarak, f ’nin tanım kümesi birkapalı aralıksa Kapalı Aralık Yöntemini kullanabilirsiniz.


Örnek 12400m çiti olan bir çiftçi, bu çit ile bir kenarı ırmağa sınır olandikdörtgensel bir arazi çevirmek istiyor. Irmak boyunca çit çekmesinegerek yoktur. En büyük alana sahip arazinin boyutları nedir?

Çözüm.

Bu problemde neler olduğunu hissetmek için birkaç özel durumudeneyelim. Şekil, 2400m’lik telin nasıl kullanılacağının olası yollarındanüçü gösterilmektedir.

3. Introduce Notation Assign a symbol to the quantity that is to be maximized orminimized (let’s call it for now). Also select symbols forother unknown quantities and label the diagram with these symbols. It mayhelp to use initials as suggestive symbols—for example, for area, forheight, for time.

4. Express in terms of some of the other symbols from Step 3.

5. If has been expressed as a function of more than one variable in Step 4, usethe given information to find relationships (in the form of equations) amongthese variables. Then use these equations to eliminate all but one of the vari-ables in the expression for . Thus, will be expressed as a function of onevariable , say, . Write the domain of this function.

6. Use the methods of Sections 4.2 and 4.3 to find the absolute maximum orminimum value of . In particular, if the domain of is a closed interval, thenthe Closed Interval Method in Section 4.2 can be used.

EXAMPLE 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular fieldthat borders a straight river. He needs no fence along the river. What are the dimen-sions of the field that has the largest area?

SOLUTION In order to get a feeling for what is happening in this problem let’s experi-ment with some special cases. Figure 1 (not to scale) shows three possible ways oflaying out the 2400 ft of fencing. We see that when we try shallow, wide fields ordeep, narrow fields, we get relatively small areas. It seems plausible that there issome intermediate configuration that produces the largest area.

Figure 2 illustrates the general case. We wish to maximize the area of the rect-angle. Let and be the depth and width of the rectangle (in feet). Then we express

in terms of and :

We want to express as a function of just one variable, so we eliminate byexpressing it in terms of . To do this we use the given information that the totallength of the fencing is 2400 ft. Thus

From this equation we have , which gives

Note that 0 and (otherwise ). So the function that we wish tomaximize is

The derivative is , so to find the critical numbers we solve the A�x� � 2400 � 4x

0 � x � 1200A�x� � 2400x � 2x 2

A � 0x � 1200x �

A � x �2400 � 2x� � 2400x � 2x 2

y � 2400 � 2x

2x � y � 2400

xyA

A � xyyxA

yxA

100

2200

100

Area=100 · 2200=220,000 ft@

700

1000

700

Area=700 · 1000=700,000 ft@

1000

400

1000

Area=1000 · 400=400,000 ft@

ff

Q � f �x�xQQ

Q

Q

thA

�a, b, c, . . . , x, y�Q

308 � CHAPTER 4 APPLICATIONS OF DIFFERENTIATION

x

y

A x

FIGURE 2

� Understand the problem� Analogy: Try special cases� Draw diagrams

� Introduce notation

FIGURE 1Alan = 220 000m2 Alan = 700 000m2 Alan = 400 000m2


Çözüm (devamı).

Sığ, geniş bölgeleri ya da derin, dar bölgeleri denediğimizde küçük alanlarelde ettiğimizi görüyoruz. En büyük alanı arada bulunan şeklin vereceğiakla yatkındır.Şekil genel durumu göstermektedir. Dikdörtgenin A alanını maksimumyapmak istiyoruz. x ve y sırasıyla dikdörtgenin derinliği ve genişliği olsun.

3. Introduce Notation Assign a symbol to the quantity that is to be maximized orminimized (let’s call it for now). Also select symbols forother unknown quantities and label the diagram with these symbols. It mayhelp to use initials as suggestive symbols—for example, for area, forheight, for time.

4. Express in terms of some of the other symbols from Step 3.

5. If has been expressed as a function of more than one variable in Step 4, usethe given information to find relationships (in the form of equations) amongthese variables. Then use these equations to eliminate all but one of the vari-ables in the expression for . Thus, will be expressed as a function of onevariable , say, . Write the domain of this function.

6. Use the methods of Sections 4.2 and 4.3 to find the absolute maximum orminimum value of . In particular, if the domain of is a closed interval, thenthe Closed Interval Method in Section 4.2 can be used.

EXAMPLE 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular fieldthat borders a straight river. He needs no fence along the river. What are the dimen-sions of the field that has the largest area?

SOLUTION In order to get a feeling for what is happening in this problem let’s experi-ment with some special cases. Figure 1 (not to scale) shows three possible ways oflaying out the 2400 ft of fencing. We see that when we try shallow, wide fields ordeep, narrow fields, we get relatively small areas. It seems plausible that there issome intermediate configuration that produces the largest area.

Figure 2 illustrates the general case. We wish to maximize the area of the rect-angle. Let and be the depth and width of the rectangle (in feet). Then we express

in terms of and :

We want to express as a function of just one variable, so we eliminate byexpressing it in terms of . To do this we use the given information that the totallength of the fencing is 2400 ft. Thus

From this equation we have , which gives

Note that 0 and (otherwise ). So the function that we wish tomaximize is

The derivative is , so to find the critical numbers we solve the A�x� � 2400 � 4x

0 � x � 1200A�x� � 2400x � 2x 2

A � 0x � 1200x �

A � x�2400 � 2x� � 2400x � 2x 2

y � 2400 � 2x

2x � y � 2400

xyA

A � xyyxA

yxA

100

2200

100

Area=100 · 2200=220,000 ft@

700

1000

700

Area=700 · 1000=700,000 ft@

1000

400

1000

Area=1000 · 400=400,000 ft@

ff

Q � f �x�xQQ

Q

Q

thA

�a, b, c, . . . , x, y�Q


x

y

A x

FIGURE 2

� Understand the problem� Analogy: Try special cases� Draw diagrams

� Introduce notation

FIGURE 1



Bu durumda, A’yı x ve y cinsinden

A = xy

şeklinde ifade ederiz.A’yı tek değişkenli fonksiyon olarak ifade etmek istiyoruz. Bu nedenley’yi x cinsinden yazarak yok edeceğiz. Bunun için, telin toplamuzunluğunun 2400m olduğu bilgisini kullanırız. Buradan,

2x+ y = 2400

olur. Bu denklemden y = 2400− 2x elde ederiz ve

A = x(2400− 2x) = 2400x− 2x2

buluruz. Bu ifadede x ≥ 0 ve x ≤ 1200 (aksi takdirde A < 0) olduğunadikkat ediniz.



ŞimdiA(x) = 2400x− 2x2 0 ≤ x ≤ 1200

fonksiyonunu maksimum yapmak istiyoruz. Türev, A′(x) = 2400− 4x’tir.Kritik sayıları bulmak için,

2400− 4x = 0

denklemini çözerek x = 600 buluruz.A’nın maksimum değeri ya bu kritik sayıda ya da aralığın bir uçnoktasında oluşur.A(0) = 0, A(600) = 720000 ve A(1200) = 0 olduğundan kapalı aralıkyöntemi maksimum değeri A(600) = 720000 olarak verir.Alternatif olarak, her x için A′′(x) = −4 < 0 olduğundan A daima içbükeydir ve x = 600’deki yerel maksimum bir mutlak maksimumolmalıdır. Sonuç olarak, dikdörtgensel bölgenin derinliği 600m vegenişliği 1200m olmalıdır.


Örnek 21 lt yağ koymak için silindir biçiminde bir teneke kutu yapılmak isteniyor.Metal maliyeti minumum olan kutu üretmek için boyutları bulunuz.

Çözüm.

Şekildeki gibi yarıçapı r’ve yüksekliği h olan bir silindir çiziniz.

equation

which gives . The maximum value of must occur either at this criticalnumber or at an endpoint of the interval. Since ,and , the Closed Interval Method gives the maximum value as

.[Alternatively, we could have observed that for all , so is

always concave downward and the local maximum at must be an absolutemaximum.]

Thus, the rectangular field should be 600 ft deep and 1200 ft wide.

EXAMPLE 2 A cylindrical can is to be made to hold 1 L of oil. Find the dimensionsthat will minimize the cost of the metal to manufacture the can.

SOLUTION Draw the diagram as in Figure 3, where is the radius and the height(both in centimeters). In order to minimize the cost of the metal, we minimize thetotal surface area of the cylinder (top, bottom, and sides). From Figure 4 we see thatthe sides are made from a rectangular sheet with dimensions and h. So thesurface area is

To eliminate we use the fact that the volume is given as 1 L, which we take tobe 1000 cm . Thus

which gives . Substitution of this into the expression for gives

Therefore, the function that we want to minimize is

To find the critical numbers, we differentiate:

Then when , so the only critical number is .Since the domain of is , we can’t use the argument of Example 1 con-

cerning endpoints. But we can observe that for and for , so is decreasing for all to the left of the critical number andincreasing for all to the right. Thus, must give rise to an absoluteminimum.

[Alternatively, we could argue that as and as ,so there must be a minimum value of , which must occur at the critical number.See Figure 5.]

The value of corresponding to is

Thus, to minimize the cost of the can, the radius should be cm and theheight should be equal to twice the radius, namely, the diameter.

s3 500��

h �1000

�r 2�

1000

��500��2�3� 2�3 500� � 2r

r � s3 500��h

A�r�r l A�r� l r l 0�A�r� l

r � s3 500��rrAr � s3 500��

A�r� � 0r � s3 500��A�r� � 0�0, �A

r � s3 500��r 3 � 500A�r� � 0

A�r� � 4�r �2000

r 2�

4��r 3 � 500�r 2

r � 0A�r� � 2�r 2 �2000

r

A � 2�r 2 � 2�r�1000�r 2 � � 2�r 2 � 2000r

Ah � 1000��r 2 �

�r 2h � 1000

3h

A � 2�r 2 � 2�rh

2�r

hr

x � 600AxA��x� � �4 � 0

A�600� � 720,000A�1200� � 0

A�0� � 0, A�600� � 720,000Ax � 600

2400 � 4x � 0

SECTION 4.6 OPTIMIZATION PROBLEMS � 309

r

h

FIGURE 3

r

Area 2{πr@}

FIGURE 4

Area (2πr)h

2πr

h

r

y

0 10

1000 y=A(r)

FIGURE 5

� In the Applied Project on page 318we investigate the most economicalshape for a can by taking into accountother manufacturing costs.

Module 4.6 takes you througheight additional optimization

problems, including animations of thephysical situations.



Metal maliyetini minimum yapmak için, silindirin toplam yüzey alanını(alt, üst ve yan) minumum yaparız. Şekilden kenarların boyutları 2πr veh olan bir dikdörtgensel levhadan yapıldığını görürüz.

equation
















s3 500��

h �1000

�r 2�

1000

��500��2�3� 2�3 500� � 2r

r � s3 500��h


r � s3 500��rrAr � s3 500��

A�r� � 0r � s3 500��A�r� � 0�0, �A

r � s3 500��r 3 � 500A�r� � 0

A�r� � 4�r �2000

r 2�

4��r 3 � 500�r 2

r � 0A�r� � 2�r 2 �2000

r

A � 2�r 2 � 2�r�1000�r 2 � � 2�r 2 � 2000r

Ah � 1000��r 2 �

�r 2h � 1000

3h

A � 2�r 2 � 2�rh

2�r

hr

x � 600AxA��x� � �4 � 0

A�600� � 720,000A�1200� � 0

A�0� � 0, A�600� � 720,000Ax � 600

2400 � 4x � 0


r

h

FIGURE 3

r

Area 2{πr@}

FIGURE 4

Area (2πr)h

2πr

h

r

y

0 10

1000 y=A(r)

FIGURE 5




equation
















s3 500��

h �1000

�r 2�

1000

��500��2�3� 2�3 500� � 2r

r � s3 500��h


r � s3 500��rrAr � s3 500��

A�r� � 0r � s3 500��A�r� � 0�0, �A

r � s3 500��r 3 � 500A�r� � 0

A�r� � 4�r �2000

r 2�

4��r 3 � 500�r 2

r � 0A�r� � 2�r 2 �2000

r

A � 2�r 2 � 2�r�1000�r 2 � � 2�r 2 � 2000r

Ah � 1000��r 2 �

�r 2h � 1000

3h

A � 2�r 2 � 2�rh

2�r

hr

x � 600AxA��x� � �4 � 0

A�600� � 720,000A�1200� � 0

A�0� � 0, A�600� � 720,000Ax � 600

2400 � 4x � 0


r

h

FIGURE 3

r

Area 2{πr@}

FIGURE 4

Area (2πr)h

2πr

h

r

y

0 10

1000 y=A(r)

FIGURE 5




Alan 2(πr2) Alan (2πr)h



Bu nedenle silindirin yüzey alanı

A = 2πr2 + 2πrh

olur. h’yi yok etmek için hacmin 1 lt olarak verildiğini 1000 cm3 alarakkullanırız. Böylece,

πr2h = 1000

den

h =1000

(πr2)

olur. Bunu A’nın ifadesinde yerine koyarsak

A = 2πr2 + 2πr

(1000

πr2

)= 2πr2 +

2000

r

elde ederiz.



Şimdi

A(r) = 2πr2 +2000

rr > 0

fonksiyonunu minimum yapmak istiyoruz. Kritik sayıları bulmak için A(r)nin türevini alırsak

A′(r) = 4πr − 2000r2

=4(πr3 − 500)

r2

olur. Burada, πr3 = 500 olduğunda A′(r) = 0 olur.

Bu nedenle, tek kritik sayı r = 3√

500π ’dir.



A’nın tanım kümesi (0,∞) olduğundan bir önceki örnekteki gibi uçnoktaları kullanamayız.

Ama r < 3√

500π için A

′(r) < 0 ve r > 3√

500π için A

′(r) > 0 olduğunu,

dolayısıyla A’nın kritik sayısının solundaki her r için azaldığını vesağındaki her r için arttığını gözlemleyebiliriz.

Böylece, r = 3√

500π mutlak minimumu vermelidir.



Alernatif olarak, r → 0+ iken A(r)→∞ ve r →∞ iken A(r)→∞olduğundan A(r)’nin bir minumum değeri olmalı ve bu minimum değerkritik sayıda meydana gelmelidir.

equation
















s3 500��

h �1000

�r 2�

1000

��500��2�3� 2�3 500� � 2r

r � s3 500��h


r � s3 500��rrAr � s3 500��

A�r� � 0r � s3 500��A�r� � 0�0, �A

r � s3 500��r 3 � 500A�r� � 0

A�r� � 4�r �2000

r 2�

4��r 3 � 500�r 2

r � 0A�r� � 2�r 2 �2000

r

A � 2�r 2 � 2�r�1000�r 2 � � 2�r 2 � 2000r

Ah � 1000��r 2 �

�r 2h � 1000

3h

A � 2�r 2 � 2�rh

2�r

hr

x � 600AxA��x� � �4 � 0

A�600� � 720,000A�1200� � 0

A�0� � 0, A�600� � 720,000Ax � 600

2400 � 4x � 0


r

h

FIGURE 3

r

Area 2{πr@}

FIGURE 4

Area (2πr)h

2πr

h

r

y

0 10

1000 y=A(r)

FIGURE 5






r = 3√

500π değerine karşılık gelen h değeri

h =1000

πr2=

1000

π(500/π)2/3= 2

3

√500

π= 2r

olur.

Böylece kutunun maliyetini minimum yapmak için yarıçap 3√

500π cm ve

yükseklik yarıçapın iki katı, yani çap olmalıdır.


Mutlak Uç Değerler için Birinci Türev Testi

c’nin, bir aralıkta tanımlı, sürekli bir f fonksiyonunun bir kritik sayısınınolduğunu varsayalım.

(a) Her x < c için f ′(x) > 0 ve her x > c için f ′(x) < 0 ise, f(c), f ’ninmutlak maksimum değeridir.

(b) Her x < c için f ′(x) < 0 ve her x > c için f ′(x) > 0 ise, f(c), f ’ninmutlak minimum değeridir.


Örnek 3y2 = 2x parabolü üzerinde (1, 4) noktasına en yakın olan noktayı bulunuz.

Çözüm.

(x, y) ve (1, 4) noktalarıarasındaki uzaklık

d =√(x− 1)2 + (y − 4)2

ile verilir.

NOTE 1 � The argument used in Example 2 to justify the absolute minimum is a vari-ant of the First Derivative Test (which applies only to local maximum or minimumvalues) and is stated here for future reference.

First Derivative Test for Absolute Extreme Values Suppose that is a critical numberof a continuous function defined on an interval.

(a) If for all and for all , then is theabsolute maximum value of .

(b) If for all and for all , then is theabsolute minimum value of .

NOTE 2 � An alternative method for solving optimization problems is to use implicitdifferentiation. Let’s look at Example 2 again to illustrate the method. We work withthe same equations

but instead of eliminating h, we differentiate both equations implicitly with respect to r :

The minimum occurs at a critical number, so we set , simplify, and arrive at theequations

and subtraction gives , or .

EXAMPLE 3 Find the point on the parabola that is closest to the point .

SOLUTION The distance between the point and the point is

(See Figure 6.) But if lies on the parabola, then , so the expressionfor becomes

(Alternatively, we could have substituted to get in terms of alone.)Instead of minimizing , we minimize its square:

(You should convince yourself that the minimum of occurs at the same point asthe minimum of , but is easier to work with.) Differentiating, we obtain

so when . Observe that when and when, so by the First Derivative Test for Absolute Extreme Values, the absolute

minimum occurs when . (Or we could simply say that because of the geomet-y � 2y � 2

f �y� � 0y � 2f �y� � 0y � 2f �y� � 0

f �y� � 2(12 y 2 � 1) y � 2�y � 4� � y 3 � 8

d 2d 2d

d 2 � f �y� � ( 12 y 2 � 1)2 � �y � 4�2

dxdy � s2x

d � s( 12 y 2 � 1)2 � �y � 4�2d

x � y 2�2�x, y�

d � s�x � 1�2 � �y � 4�2

�x, y��1, 4�

�1, 4�y 2 � 2x

h � 2r2r � h � 0

2h � rh � 02r � h � rh � 0

A � 0

2�rh � �r 2h � 0A � 4�r � 2�h � 2�rh

�r 2h � 100A � 2�r 2 � 2�rh

ff �c�x � cf �x� � 0x � cf �x� � 0

ff �c�x � cf �x� � 0x � cf �x� � 0

fc


x

y

0 1

1

2 3 4

¥=2x(1, 4)

(x, y)

FIGURE 6



Ama, (x, y) parabolün üzerindeyse x = 12y2’dir ve d’nin ifadesi

d =

√(1

2y2 − 1

)2+ (y − 4)2

olur (ikinci seçenek, y =√2x alarak d’yi yalnızca x cinsinden ifade

edebilirdik).d’yi minimum yapmak yerine karesi olan

d2 = f(y) =

(1

2y2 − 1

)2+ (y − 4)2

fonksiyonunu minimum yapacağız. d2’nin minimumu ile d’ninminimumunun aynı noktada meydana geldiğine ama d2 ile çalışmanın dile çalışmaktan daha kolay olduğuna dikkat ediniz.



Türev alırsak

f ′(y) = 2

(1

2y2 − 1

)y + 2(y − 4) = y3 − 8

elde ederiz. Dolayısıyla, y = 2 olduğunda f ′(y) = 0 olur.y < 2 için f ′(y) < 0 ve y > 2 için f ′(y) > 0 olduğunu gözlemleyiniz.Buradan mutlak uç değerler için birinci türev testine göre mutlakminimum y = 2’de meydana gelir.(Yalnızca, problemin geometrik yapısından ötürü en yakın noktanınolduğunun fakat en uzak noktanın olmadığının açık olduğunusöyleyebilirdik.) Karşı gelen x değeri x = 12y

2 = 2’dir.Böylece, y2 = 2x parabolü üzerinde (1, 4) noktasına en yakın olan nokta,(2, 2) noktasıdır.


Örnek 4r yarıçaplı bir yarıçember içine çizilebilecek en büyük dikdörtgenin alanınıbulunuz.

Çözüm.

Yarıçember olarak, merkezi başlangıç noktası olan x2 + y2 = r2

çemberinin üst yarısını alalım. Bu durumda, yarıçemberin içine çizmekdemek, Şekilde gösterildiği gibi dikdörtgenin iki köşesinin yarıçemberüzerinde, iki köşesinin de x-ekseni üzerinde olması demektir.

EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircleof radius .

SOLUTION 1 Let’s take the semicircle to be the upper half of the circle with center the origin. Then the word inscribed means that the rectangle has twovertices on the semicircle and two vertices on the -axis as shown in Figure 9.

Let be the vertex that lies in the first quadrant. Then the rectangle has sidesof lengths and , so its area is

To eliminate we use the fact that lies on the circle and so. Thus

The domain of this function is . Its derivative is

which is 0 when , that is, (since ). This value of gives a maximum value of since and . Therefore, the area of the largestinscribed rectangle is

SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable.Let be the angle shown in Figure 10. Then the area of the rectangle is

We know that has a maximum value of 1 and it occurs when . Sohas a maximum value of and it occurs when .

Notice that this trigonometric solution doesn’t involve differentiation. In fact, we didn’t need to use calculus at all.

� � ��4r 2A��2� � ��2sin 2�

A�� 2r cos ��r sin �� r 2�2 sin � cos �� r 2 sin 2�

�

A� rs2� � 2 rs2�r 2 � r 22 � r 2

A�r� � 0A�0� � 0Axx � 0x � r�s22x 2 � r 2

A � 2sr 2 � x 2 �2x 2

sr 2 � x 2�

2�r 2 � 2x 2 �sr 2 � x 2

0 � x � r

A � 2xsr 2 � x 2y � sr 2 � x 2

x 2 � y 2 � r 2�x, y�y

A � 2xy

y2x�x, y�

x

x 2 � y 2 � r 2

r


always 23. On the basis of the evidence in your table,estimate the answer to the problem.

(b) Use calculus to solve the problem and compare withyour answer to part (a).

2. Find two numbers whose difference is 100 and whose prod-uct is a minimum.

3. Find two positive numbers whose product is 100 and whosesum is a minimum.

4. Find a positive number such that the sum of the number andits reciprocal is as small as possible.

1. Consider the following problem: Find two numbers whosesum is 23 and whose product is a maximum.(a) Make a table of values, like the following one, so that

the sum of the numbers in the first two columns is

Exercises � � � � � � � � � � � � � � � � � � � � � � � � � �4.6

First number Second number Product

1 22 222 21 423 20 60. . .. . .. . .

x

y

0

2x

(x, y)

y

_r r

FIGURE 9

Resources / Module 5/ Max and Min

/ Start of Max and Min

r

¨

r Ł ¨

r N ¨

FIGURE 10



(x, y) noktası birinci bölgedeki köşe olsun. Bu durumda dikdörtgeninkenasr uzunlukları 2x ve y’dir. Dolayısıyla, dikdörtgenin alanı

A = 2xy

olur. y’yi yok etmek için (x, y) noktasının x2 + y2 = r2 çemberi üzerindeolduğunu kullanarak y =

√r2 − x2 yazarız. Böylece,

A = 2x√r2 − x2

olur. Bu fonksiyonun tanım kümesi 0 ≤ x ≤ r’dir.



Türevi,

A′ = 2√r2 − x2 − 2x

2

√r2 − x2

=2(r2 − 2x2)√

r2 − x2

olarak bulunur ve bu türev 2x2 = r2 olduğunda 0’dır. Buradan, (x ≥ 0olduğundan) x = r√

2buluruz. A(0) = 0 ve A(r) = 0 olduğundan x’in bu

değeri A’nın maksimum değerini verir. Dolayısıyla, yarı çember içineçizilebilecek en büyük dikdörtgenin alanı

A

(r√2

)= 2

r√2

√r2 − r

2

2= r2

olur.

Documents

Optimizasyon Problemlerinin C˘ozum unde Takip Edilecek Ad mlarkisi.deu.edu.tr/.../optimizasyonproblemleri.pdf · Optimizasyon Problemleri 4 Q’yu Ad m3’deki di ger semboller cinsinden