Maxima and Minima in Plane and Solid Figures

Lesson 8-3

Optimization

• Finding the maximum/minimum (as in the previous lesson) is an important part of problem solving whether in relation to maximizing profit, minimizing cost in manufacturing, of maximizing volume (to mention a few applications).

• The process of maximizing or minimizing is called optimization.

Optimization Guidelines

1) Read and understand the problem. Identify the given quantities and those you must find.

2) Sketch a diagram and label it appropriately, introducing variables for unknown quantities.

3) Decide which quantity is to be optimized and express this quantity as a function f of one or more other variables.

Optimization Guidelines…

4) Using available information, express f as a function of just one variable.

5) Determine the domain of f and draw its graph.

6) Find the global extrema of f, considering any critical points and endpoints.

7) Convert the results obtained on step 6 back into the context of the original problem. Be sure you have answered the question originally asked.

Example 1:

An open box with a rectangular base is to be constructed from a rectangular piece of cardboard 16 inches wide and 21 inches long by cutting congruent squares from each corner and then bending up the sides. Find the size of a corner square that will produce an open-top box with the largest possible volume.

Example 1: Step 2, 3 and 416

21

x

x21

– 2x

16 – 2x

V = LWH

V = (16 – 2x)(21 – 2x)(x) = 4x3 – 74x2 + 336x

Domain of V is 0 < x < 8

Example 1: Step 5

Domain of V is 0≤ x ≤ 8

Window

x[0, 9]

y[0, 500]

yscl 100

V = 4x3 – 74x2 + 336x

Example 1: Step 6

212 148 336dV

x xdx

24(3 37 84)x x

24(3 37 84) 0x x

Recall, critical numbers exist where the derivative is zero or does not exist!!!!

4( 3)(3 28) 0x x

3 0x 3 28 0x 3x 1

39x

Outside the

domain!

x = 0 or 8 gives no volume

Example 1: Step 7

x21

– 2x

16 – x x = 3

V(3) = 4(3)3 – 74(3)2 + 336(3)

The volume is maximized at 450 in3 when the corner square is 3 in. x 3 in.

Answer the original question!!!

= 450

Example 2:

10”

Step 2 Draw and label a diagram.

Step 1 Read and understand the problem

6”

Find the radius and height of the right-circular cylinder of largest volume that can be inscribed in a right-circular cone with radius 6 in. and height 10 in.

h

r

Use similar triangles to get h in terms of r.

Step 4 Express V as a function of one

variable.

Example 2:

10 60 6r h

10 106

hr

53 10h r

10”

V = πr2h

6”

Step 3 and 4

h

r

Step 3 Quantity to be optimized.

r

6

h10”

10–h

2 53 10V r r

3 253 10r rNote, had we put r in

terms of h we would have had to square it.

Example 2:

10”

Domain of V is 0 < r < 6

The radius of the cylinder can not be

greater than the cone…6

6”

h

r

Step 5 Step 5 Determine

the domain and graph.

3 253 10V r r

Example 2: Step 6 and 7

25 20dV

r rdr

2( 5 20 ) 0r r

5 ( 4) 0r r

Recall, critical numbers exist where the derivative is zero or does not exist!!!!

0r 4r

203

@ 4 10r h

53

10h r

Therefore, the inscribed cone of largest volume has a radius of 4 in. and height of 3 1/3 in.

13

3

Example 3:

Step 2 Draw and label a diagram.

A rectangle is inscribed between the graphs of y = ¼ x4 -1 and

y = 4-x2. Find the width of the rectanglethat has the largest area.

Step 1 Read and understand the problem (x1, y1)

(x1, y2)

(x2, y1)

Example 3: 1 2Area = 2 ( )x yy

2 414(2 ( )4 1)x xA x

4124 1))2 4( xA x x

Step 3 and 4

Area = L • W

(x1, y1)

(x1, y2)

(x2, y1)

3 5128 2 2A x x x x

5 312 2 10A x x x

2 4144 1x x

4 214 5 0x x

Example 3: Step 5 and 6

Using solve on the TI-89 yields 2 6 2 1.703x

4 256 10

2dA

x xdx

4 256 10 0

2x x

10 ( 34 3)5

x 1.064

Example 3: Step 7

Ω

1.064xTherefore, the width of 2(1.064) or about 2.128 will yield the largest area of the rectangle between the curves.