It is not a coincidence!On patterns in some Calculus

optimization problems.

Maria NoginCalifornia State University, Fresno

mnogin@csufresno.edu

Outline

1 The basicsOptimizing rectangleOptimizing rectangular box

2 The rectangular field problemProblemObservationWhy?

3 The can problemProblemObservationWhy?

Optimizing rectangle

Out of all rectangles with a given perimeter, which one has thegreatest area?

s

h h

s−h

s+h

s

s−h

h h

s−h

s+h

Optimizing rectangle

Out of all rectangles with a given perimeter, which one has thegreatest area?

s

h h

s−h

s+h

s

s−h

h h

s−h

s+h

Optimizing rectangle

Out of all rectangles with a given perimeter, which one has thegreatest area?

s

h h

s−h

s+h

s

s−h

h h

s−h

s+h

Optimizing rectangle

Out of all rectangles with a given perimeter, which one has thegreatest area?

s

h h

s−h

s+h

s

s−h

h h

s−h

s+h

Optimizing rectangular box

Out of all rectangular boxes with a given volume, which one hasthe smallest surface area?

Optimizing rectangular box

Out of all rectangular boxes with a given volume, which one hasthe smallest surface area?

Optimizing rectangular box

Out of all rectangular boxes with a given volume, which one hasthe smallest surface area?

Optimizing rectangular box

Out of all rectangular boxes with a given volume, which one hasthe smallest surface area?

Optimizing rectangular box

Out of all rectangular boxes with a given volume, which one hasthe smallest surface area?

The rectangular field roblem

A farmer wants to fence off a rectangular field and divide it into 3pens with fence parallel to one pair of sides. He has a total 2400 ftof fencing. What are the dimensions of the field has the largestpossible area?

y

x

y = 2400−4x2 = 1200 − 2x

Area(x) = 1200x − 2x2

Area′(x) = 1200 − 4x2 = 0

x = 300 is an absolute maximum

y = 600

Observation: the total length of vertical pieces: 1200 ftthe total length of horizontal pieces: 1200 ft

These are equal!

The rectangular field roblem

A farmer wants to fence off a rectangular field and divide it into 3pens with fence parallel to one pair of sides. He has a total 2400 ftof fencing. What are the dimensions of the field has the largestpossible area?

y

x

y = 2400−4x2 = 1200 − 2x

Area(x) = 1200x − 2x2

Area′(x) = 1200 − 4x2 = 0

x = 300 is an absolute maximum

y = 600

Observation: the total length of vertical pieces: 1200 ftthe total length of horizontal pieces: 1200 ft

These are equal!

The rectangular field roblem

A farmer wants to fence off a rectangular field and divide it into 3pens with fence parallel to one pair of sides. He has a total 2400 ftof fencing. What are the dimensions of the field has the largestpossible area?

y

x

y = 2400−4x2 = 1200 − 2x

Area(x) = 1200x − 2x2

Area′(x) = 1200 − 4x2 = 0

x = 300 is an absolute maximum

y = 600

Observation: the total length of vertical pieces: 1200 ftthe total length of horizontal pieces: 1200 ft

These are equal!

The rectangular field roblem

A farmer wants to fence off a rectangular field and divide it into 3pens with fence parallel to one pair of sides. He has a total 2400 ftof fencing. What are the dimensions of the field has the largestpossible area?

y

x

y = 2400−4x2 = 1200 − 2x

Area(x) = 1200x − 2x2

Area′(x) = 1200 − 4x2 = 0

x = 300 is an absolute maximum

y = 600

Observation: the total length of vertical pieces: 1200 ftthe total length of horizontal pieces: 1200 ft

These are equal!

The rectangular field roblem

A farmer wants to fence off a rectangular field and divide it into 3pens with fence parallel to one pair of sides. He has a total 2400 ftof fencing. What are the dimensions of the field has the largestpossible area?

y

x

y = 2400−4x2 = 1200 − 2x

Area(x) = 1200x − 2x2

Area′(x) = 1200 − 4x2 = 0

x = 300 is an absolute maximum

y = 600

Observation: the total length of vertical pieces: 1200 ftthe total length of horizontal pieces: 1200 ft

These are equal!

Why? Functional explanation

y

x

Let L be the total length of the vertical pieces.2400 − L is the total length of the horizontal pieces.

x = L4 , y = 2400−L

2 , Area(L) = L4 · 2400−L

2

24001200

Area

L0

Why? Functional explanation

y

x

Let L be the total length of the vertical pieces.

2400 − L is the total length of the horizontal pieces.

x = L4 , y = 2400−L

2 , Area(L) = L4 · 2400−L

2

24001200

Area

L0

Why? Functional explanation

y

x

Let L be the total length of the vertical pieces.2400 − L is the total length of the horizontal pieces.

x = L4 , y = 2400−L

2 , Area(L) = L4 · 2400−L

2

24001200

Area

L0

Why? Functional explanation

y

x

Let L be the total length of the vertical pieces.2400 − L is the total length of the horizontal pieces.

x = L4 ,

y = 2400−L2 , Area(L) = L

4 · 2400−L2

24001200

Area

L0

Why? Functional explanation

y

x

Let L be the total length of the vertical pieces.2400 − L is the total length of the horizontal pieces.

x = L4 , y = 2400−L

2 ,

Area(L) = L4 · 2400−L

2

24001200

Area

L0

Why? Functional explanation

y

x

Let L be the total length of the vertical pieces.2400 − L is the total length of the horizontal pieces.

x = L4 , y = 2400−L

2 , Area(L) = L4 · 2400−L

2

24001200

Area

L0

Why? Functional explanation

y

x

Let L be the total length of the vertical pieces.2400 − L is the total length of the horizontal pieces.

x = L4 , y = 2400−L

2 , Area(L) = L4 · 2400−L

2

24001200

Area

L0

Why? Geometric explanation

x

y

y

x

x

Why? Geometric explanation

x

y y

x

x

Why? Geometric explanation

x

y

y

x

x

2x/

Why? Geometric explanation

x

y y

x

x

2x/

The can problem

A cylindrical can has to have volume 1000cm3. Find thedimensions of the can that minimize the amount of material used(i.e. minimize the surface area).

h

r

h = 1000πr2

SA(r) = 2πr2 + 2000r

SA′(r) = 4πr − 2000r2

= 0

r = 3

√500π is an absolute minimum

h = 2 3

√500π

Observation: d = 2 3

√500π cm

d = h!

The can problem

A cylindrical can has to have volume 1000cm3. Find thedimensions of the can that minimize the amount of material used(i.e. minimize the surface area).

h

r

h = 1000πr2

SA(r) = 2πr2 + 2000r

SA′(r) = 4πr − 2000r2

= 0

r = 3

√500π is an absolute minimum

h = 2 3

√500π

Observation: d = 2 3

√500π cm

d = h!

The can problem

A cylindrical can has to have volume 1000cm3. Find thedimensions of the can that minimize the amount of material used(i.e. minimize the surface area).

h

r

h = 1000πr2

SA(r) = 2πr2 + 2000r

SA′(r) = 4πr − 2000r2

= 0

r = 3

√500π is an absolute minimum

h = 2 3

√500π

Observation: d = 2 3

√500π cm

d = h!

The can problem

A cylindrical can has to have volume 1000cm3. Find thedimensions of the can that minimize the amount of material used(i.e. minimize the surface area).

h

r

h = 1000πr2

SA(r) = 2πr2 + 2000r

SA′(r) = 4πr − 2000r2

= 0

r = 3

√500π is an absolute minimum

h = 2 3

√500π

Observation: d = 2 3

√500π cm

d = h!

The can problem

A cylindrical can has to have volume 1000cm3. Find thedimensions of the can that minimize the amount of material used(i.e. minimize the surface area).

h

r

h = 1000πr2

SA(r) = 2πr2 + 2000r

SA′(r) = 4πr − 2000r2

= 0

r = 3

√500π is an absolute minimum

h = 2 3

√500π

Observation: d = 2 3

√500π cm d = h!

Why?

h

r

h

r

Vcan = Acircleh Vcube = Asquareh

Vcube =Asquare

AcircleVcan = 4

πVcan

SAcan = 2Acircle + Pcircleh SAcube = 2Asquare + Psquareh

Question: isPsquare

Pcircle=

Asquare

Acircle?

Answer: 8r2πr = 4r2

πr2Yes!

Why?

h

r

h

r

Vcan = Acircleh Vcube = Asquareh

Vcube =Asquare

AcircleVcan = 4

πVcan

SAcan = 2Acircle + Pcircleh SAcube = 2Asquare + Psquareh

Question: isPsquare

Pcircle=

Asquare

Acircle?

Answer: 8r2πr = 4r2

πr2Yes!

Why?

h

r

h

r

Vcan = Acircleh Vcube = Asquareh

Vcube =Asquare

AcircleVcan = 4

πVcan

SAcan = 2Acircle + Pcircleh SAcube = 2Asquare + Psquareh

Question: isPsquare

Pcircle=

Asquare

Acircle?

Answer: 8r2πr = 4r2

πr2Yes!

Why?

h

r

h

r

Vcan = Acircleh Vcube = Asquareh

Vcube =Asquare

AcircleVcan = 4

πVcan

SAcan = 2Acircle + Pcircleh SAcube = 2Asquare + Psquareh

Question: isPsquare

Pcircle=

Asquare

Acircle?

Answer: 8r2πr = 4r2

πr2Yes!

Why?

h

r

h

r

Vcan = Acircleh Vcube = Asquareh

Vcube =Asquare

AcircleVcan = 4

πVcan

SAcan = 2Acircle + Pcircleh SAcube = 2Asquare + Psquareh

Question: isPsquare

Pcircle=

Asquare

Acircle?

Answer: 8r2πr = 4r2

πr2Yes!

Why?

h

r

h

r

Vcan = Acircleh Vcube = Asquareh

Vcube =Asquare

AcircleVcan = 4

πVcan

SAcan = 2Acircle + Pcircleh SAcube = 2Asquare + Psquareh

Question: isPsquare

Pcircle=

Asquare

Acircle?

Answer: 8r2πr = 4r2

πr2Yes!

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare=

Ahexagon

Phexagon

πr2

2πr = 4r2

8r = ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare=

Ahexagon

Phexagon

πr2

2πr = 4r2

8r = ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare=

Ahexagon

Phexagon

πr2

2πr = 4r2

8r = ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare=

Ahexagon

Phexagon

πr2

2πr = 4r2

8r = ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare

=Ahexagon

Phexagon

πr2

2πr = 4r2

8r

= ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare

=Ahexagon

Phexagon

πr2

2πr = 4r2

8r

= ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br

= cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare

=Ahexagon

Phexagon

πr2

2πr = 4r2

8r

= ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br

= cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare

=Ahexagon

Phexagon

πr2

2πr = 4r2

8r

= ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare=

Ahexagon

Phexagon

πr2

2πr = 4r2

8r

= ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Is that a coincidence?

WhyPsquare

Pcircle=

Asquare

Acircle?

Equivalently:

AcirclePcircle

=Asquare

Psquare

πr2

2πr = 4r2

8r

AcirclePcircle

=Asquare

Psquare=

Ahexagon

Phexagon

πr2

2πr = 4r2

8r = ??

The denominator is the derivative of the numerator!

r

∆ r

r

∆ r

r

∆ r

ar2

2ar = br2

2br = cr2

2cr

Other boxes

h

r

h

r r

h

Optimal shape: h = 2r

Thank you!