4.6 Optimization

• The goal is to maximize or minimize a given quantity subject to a constraint.

• Must identify the quantity to be optimized – along with the appropriate equation.

• Must identify the constraint – generally the constraint will be able to be expressed as an equation.

• Express the quantity to be optimized as a function of one variable using substitution from the constraint.

• Determine a domain that makes sense for the function to be optimized.

• Find the derivative of the function to be optimized.

• Test the derivative for extrema (a.k.a. find critical values…)

• Determine which extrema answer the question and calculate max/min values as appropriate.

ex: We need to enclose a field with a fence.  We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing.  Determine the dimensions of the field that will enclose the largest area.

Draw:

In this problem we want to maximize the area of a field and we know that we will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are:

Maximize: A = xyConstraint: 500 = x + 2y

We want to write Area in terms of one variable – so solve for one variable using the equation for the constraint and sub it into the area formula…

500 – 2y = x A = (500-2y) y = 500y – 2y2

Now we want to find the largest value this will have on the interval [0,250].  Note that the interval corresponds to taking y = 0 (i.e. no sides to the fence) and y = 250  (i.e. only two sides and no width - if there are two sides each must be 250 ft to use the whole 500ft…). 

Find the derivative of the area function:

A(y) = 500y – 2y2 A’(y) = 500 – 4y

Find critical points for the derivative:

500 – 4y = 0500 = 4yy = 125

This y-value produces an area of 31,250 ft2 and is the largest possible area.The dimensions of the fencing will be:

500 = x + 2y x = 250, y = 125

Building

125

250

HW 4.6A - pg 316#1

HW Quiz 4.6#1, (b) thru (f)

ex: A manufacturer needs to make a cylindrical can that will hold 1.5 liters of liquid.  Determine the dimensions of the can that will minimize the amount of material used in its construction.

Minimize: Surface Area

Constraint: V = 1.5L

Formula’s:• Think of volume for a cylinder as the area of a bunch of circles stacked up on one another – the height of the stack of circles is the height of the cylinder – so for a cylinder:

hrV 2π=

Surface Area for a cylinder is just the circumference of a circle in the stack, times the height of the stack PLUS the areas of the two circles making up the top and bottom of the cylinder:

222 rrhA ππ +=

• Convert liters to units of length: 1 Liter = 1000 cm3 so 1.5 liters = 1500cm3.This will then give a radius and height in terms of centimeters.

• In this case it looks like our best option is to solve the constraint for h and plug this into the area function:

• Find the derivative of this area:

2

3

2

2

3000430004)('

30002)(

r

r

rrrA

rrrA

−=⎟

⎠

⎞⎜⎝

⎛−=

⎟⎠

⎞⎜⎝

⎛+=

ππ

π

Now locate critical values for this derivative

2

3 30004)('

r

rrA

−=

π0

030004 3

=→=−→

rrπ

2035.6750

4

3000

4

3000

30004

030004

33

3

3

3

===

=

=

=−

ππ

π

ππ

r

r

r

r

                                                    Determine height of the can…

Therefore if the manufacturer makes the can with a radius of 6.2035 cm and a height of 12.4070 cm, the least amount of material will be used to make the can.

ex: We want to construct a box with a square base and we only have 10 m2 of material to use in the construction of the box.  Assuming that all the material is used in the construction process, determine the maximum volume that the box can have.

• Maximize: Volume• Constraint: Surface Area = 10

Equationsl

w

h

l = w

10)2(21042 2 =+→=+ hwwwhw

hwhwlV 2=••=

Solve constraint for h…

w

www

ww

wwh

ww

h

whw

2

5

2

5

2

5

2210

2

102

2

10)2(

2

2

−=

−

=−

=−

=

−=

=+

**2nd derivative determines concavity… a negative 2nd derivative means the function is always CC down, thus any critical number is an automatic global max.

hwV 2=

w

wh

2

5 2−=

€

V = w2 5 −w2

2w

⎛

⎝ ⎜

⎞

⎠ ⎟= w

5 −w2

2

⎛

⎝ ⎜

⎞

⎠ ⎟=

1

25w −w3

( )

Substitute…

Calculate derivative…

€

V =1

25w −w3

( )

dV

dw=

1

25 − 3w2

( )

d2V

dw2= −3w

Solve for critical numbers…

1.2913

5

3

5

53

035

2

2

2

==

=

−=−

=−

w

w

w

w

Substitute for maximum volume…

€

V =1

25w −w3

( )

V =1

25 1.291( ) − 1.291( )

3

( )

V = 2.1517m3

ex: Determine the point(s) on y = x2 + 1 that are closest to (0,2)

Key Equation: The Distance Formula:

Minimize: DistanceConstraint: y = x2 + 1

• We need to minimize the distance between the point (0,2) and any point that is on the graph (x,y).

• Equation:

€

d = x − 0( )2

+ y − 2( )2

= x 2 + y − 2( )2

** The point that minimizes the distance will also minimize the square of the distance. So since it will be easier to work with, we will use the square of the distance and minimize that:

€

d2 = x 2 + y − 2( )2

• Compare this with the constraint now… y = x2 + 1• Which variable would be easier to plug into the distance formula?

Positive second derivative means any critical number is an automatic MIN.

( )222 2−+= yxd12 −=yx

Substitute…

€

d2 = y −1+ y − 2( )2

d2 = y −1+ y 2 − 4y + 4

d2 = y 2 − 3y + 3

Calculate derivative…

2

32

33

22

2

2

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

+−=

ddyd

yddyd

yyd

Solve for critical numbers…

2

3

32

032

=

==−

y

yy

So all that we need to do at this point is find the value(s) of x that go with this value of y:

2

1

2

1

12

3

1

2

2

2

±=

=

−=

−=

x

x

x

yx

Our two closest points are then:

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛

2

3,

2

1

2

3,

2

1and

HW 4.6B – pg 316

#’s 2-4 all, 7, 8

HW 4.6B – pg 316

#’s 2-4 all, 7, 8

ex: Determine the area of the largest rectangle that can be inscribed in a circle of radius 4...

• We want the area of the largest rectangle that we can fit inside a circle and have all of its corners touching the circle.

• To do this problem it’s easiest to assume that the circle (and hence the rectangle) is centered at the origin.  Doing this we know that the equation of the circle will be:

1622 =+ yxand that the right upper corner of the rectangle will have the coordinates (x, y).

xyyxA 422 =•=

Constraint? 1622 =+ yx

xyA 4=1622 =+ yx

Solve constraint for y…

2

22

16

16

xy

xy

−=

−=Substitute…

2164 xxA −=

Calculate derivative…

€

A = 4x 16 − x 2

A'= 4 16 − x 2 + 4x1

216 − x 2

( )−1/ 2

• −2x ⎡ ⎣ ⎢

⎤ ⎦ ⎥

= 4 16 − x 2 + 4x−x

16 − x 2

⎡

⎣ ⎢

⎤

⎦ ⎥

= 4 16 − x 2 −4x 2

16 − x 2=

64 − 4x 2( ) − 4x 2

16 − x 2=

64 − 8x 2

16 − x 2

Limits? {0 < x < 4}

Calculate critical numbers…

016

0864

16

864'

2

2

2

2

=−→

=−→

−

−=

x

x

x

xA

22

8

648

0864

2

2

2

=

=

=

=−

x

x

x

x

4

16

016

016

2

2

2

==

=−

=−

xx

x

x

Calculate y-coordinates for critical numbers…

216 xy −= 04164 2 =−=→= yx

€

x = 2 2 → y = 16 − 2 2( )2

= 16 − 8 = 8 = 2 2

• It looks like the maximum area will be found if the inscribed rectangle is in fact a square:

€

2 2, 2 2( )

€

A = 4xy

A = 4 • 2 2( ) • 2 2( )

A = 4 • 8 = 32

HW 4.6C pg 316

#’s 10-12 all