maxima and minima in plane and solid figures lesson 8-3
TRANSCRIPT
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.