Transcript Optimization - robeson.k12.nc.us

Optimization
Applications of Differentiation
What is optimization?
Some of the most important applications of
differential calculus are optimization
problems, in which we are required to find the
optimal (best) way of doing something. Many
practical problems require us to minimize a
cost or maximize an area or somehow find the
best possible outcome of a situation.
Quick Review of
Helpful Formulas
A = πr²
C = 2πr
V = πr²h
SA = 2πr² + 2πrh
(closed top)
SA = πr² + 2πrh
(open top)
Steps in Solving
Optimization Problems
1.
2.
3.
4.
Read the problem.
Draw and label a picture with the given information.
Write 1 equation for every variable.
Solve for either x or y in the equation containing a
numerical value.
5. Substitute x or y into the other equation.
6. Take the derivative.
7. Set the derivative equal to 0, and solve for either x or y.
8. Take the second derivative to determine whether the
answer is a minimum or maximum value.
(< 0 = MAX, > 0 = MIN)
9. Using the original equation, solve for the remaining
variable.
Ex. 1
READ THE PROBLEM:
A rectangular field is to be fenced off along the
bank of a river, and no fence is required along the
river. The material for the fence costs $4 per running
foot for the two ends and $6 per running foot for
the side parallel to the river. Find the dimensions of
the field of largest possible area that can be
enclosed with $1,800 worth of fence.
x = $4
WRITE 1 EQUATION FOR
EVERY VARIABLE:
y = $6
x
A = xy
C = 2(4x) + 6y
1,800 = 8x + 6y
Ex. 1
SOLVE FOR EITHER X OR Y IN
THE EQUATION CONTAINING
A NUMERICAL VALUE :
1,800 = 8x + 6y
900 = 4x + 3y
900 – 3y = 4x
225 – ¾y = x
x = $4
y = $6
x
SUBSTITUTE EITHER X OR Y INTO
THE OTHER EQUATION:
A = xy
A = (225 – ¾y)y
A = 225y – ¾y²
Ex. 1
TAKE THE DERIVATIVE, SET EQUAL TO
ZERO, AND SOLVE FOR EITHER X OR Y:
A = 225y – ¾y²
A’ = 225 – (3/2)y
0 = 225 – (3/2)y
225 = (3/2)y
150 = y
x = $4
y = $6
x
TAKE THE 2ND DERIVATIVE TO
DETERMINE WHETHER THE
ANSWER IS A MIN OR MAX:
A’’ = -3/2
-3/2 < 0
MAX
In this case, the answer is
a maximum value
because we were asked
to find the dimensions of
the field of LARGEST
possible area.
Ex. 1
USING THE ORIGINAL EQUATION,
SOLVE FOR THE REMAINING VARIABLE:
y = 150
x = 225 – (3/4)y
x = 225 – (3/4)(150)
x = 225/2 = 112.5
x = $4
FINAL ANSWER
(DIMENSIONS OF THE FIELD):
y = $6
x
x = 112.5 ft
y = 150 ft
Ex. 2
x
16 in.
x
READ THE PROBLEM:
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 out a
square from each corner and
then bending up the sides. Find
the size of the corner square
which will produce a box
having the largest possible
volume.
23 in.
WRITE 1 EQUATION FOR
EVERY VARIABLE:
V = lwh
V = x (21-2x) (16-2x)
V = 336x – 74x² + 4x³
Ex. 2
x
16 in.
x
TAKE THE DERIVATIVE, SET
EQUAL TO ZERO, AND
SOLVE FOR EITHER X OR Y:
V = 336x – 74x² + 4x³
V’ = 336 – 148x + 12x²
0 = 4 (3x² - 37x + 84)
0 = 4 (3x – 28) (x – 3)
x = 28/3, x = 3
23 in.
In this case, x cannot
equal 28/3 because the
length of one side of the
paper is only 16 inches.
28/2 multiplied by 2
would be 56/3 or
roughly 18.7 inches.
Ex. 2
16 in.
x
x
2ND DERIVATIVE:
V’ = 336 – 148x + 12x²
V’’ = 24x – 148
V’’ < 0
MAX
23 in.
FINAL ANSWER
(SIZE OF CORNER SQUARE):
x² = 9 in.²
Ex. 3
READ THE PROBLEM:
An oil can is to be made in the form of a right
circular cylinder to contain K cubic centimeters.
What dimensions of the can will require the
least amount of material?
WRITE 1 EQUATION FOR EVERY
VARIABLE, AND SOLVE FOR h IN
THE EQUATION CONTAINING A
NUMERICAL VALUE:
V = K cm³
K = πr²h
h = K / πr²
SA = 2πr² + 2πrh
Ex. 3
SUBSTITUTE h INTO THE OTHER EQUATION:
h = K / πr²
SA = 2πr² + 2πr(K / πr²)
SA = 2πr² + (2K / r)
TAKE THE DERIVATIVE, SET EQUAL
TO ZERO, AND SOLVE FOR r:
SA’ = 4πr – (2K / r²)
4πr = 2K / r²
4πr³ = 2K
r³ = 2K / 4π
r = (K / 2π) 1/3
2ND DERIVATIVE:
SA’’ = 4π + (4K / r³)
SA’’ > 0
MIN
Ex. 3
USING THE ORIGINAL EQUATION,
SOLVE FOR THE REMAINING VARIABLE:
r = (K / 2π) 1/3
h = K / πr²
h = K / π(K/2π)2/3 • (K / 2π) 1/3 / (K / 2π) 1/3
h = 2 (K / 2π) 1/3
FINAL ANSWER
(DIMENSIONS OF THE CAN):
r = (K / 2π) 1/3
h = 2 (K / 2π) 1/3
Now it’s time
to try some
on your own!
Try Me #1
A window is in the
shape of a
rectangle
surmounted by a
semicircle. Find the
dimensions when
the perimeter is 12
meters and the
area is as large as
possible.
Sol. 1
First, lets draw our picture…
X
Y
2X
Sol. 1
P = 2x + 2y + πx
12 = 2x + 2y + πx
y = (12 – 2x – πx) / 2
y = 6 – x – (1/2)πx
X
2X
A = 2xy + (1/2)πx²
A = 2x [6 – x – (1/2)πx] + (1/2)πx²
A = 12x – 2x² - πx² + (1/2)πx²
A = 12x – 2x² - (1/2)πx²
Y
A’ = 12 – 4x – πx
12 = x (4 + π)
X = 12 / (4 + π)
Sol. 1
X = 12 / (4 + π)
A’ = 12 – 4x – πx
A’’ = -4 – π
A’’ < 0
MAX
X
y = 6 – x – (1/2)πx
y = 6 – [12 / (4 + π)] – (1/2)π[12 / (4 + π)]
y = 24 / (4 + π)
Y
2X
Final Answer
(Dimensions of the Window):
x = 12 / (4 + π)
y = 24 / (4 + π)
Try Me #2
A circular cylindrical
container, open at the top
and having a capacity of 24π
cubic inches, is to be
manufactured. If the cost of
the material used for the
bottom of the container is
three times that used for the
curved part, and if there is
no waste of material, find
the dimensions which will
minimize the cost.
Sol. 2
First, lets draw our picture…
r
h
Sol. 2
V = 24π
V = πr²h
24π = πr²h
h = 24 / r²
r
h
SA = 3(πr²) + 1(2πrh)
SA = 3πr² + 2πr (24 / r²)
SA = 3πr² + (48π / r)
SA’ = (-48π / r²) + 6πr
6πr = 48π / r²
6πr³ = 48π
r=2
Sol. 2
r=2
SA’ = (-48π / r²) + 6πr
SA’’ = (96π / r³) + 6π
SA’’ > 0
MIN
r
h
h = 24 / r²
h = (24 / 4) = 6
Final Answer
(Dimensions of the Can):
r=2
h=6
Try Me #3
A right circular cylinder is
inscribed in a right circular
cone so that the centerlines
of the cylinder and the cone
coincide. The cone has a
height of 6 and a radius of
base 3. Find the volume and
the dimensions of the
cylinder that has maximum
volume.
Sol. 3
First, lets draw our picture…
6
h
3
Sol. 3
6 / 3 = (6 – h) / r
h = 6 – 2r
r
h
3
6
V = πr²h
V = πr² (6 – 2r )
V = 6πr² - 2πr³
V’ = 12πr – 6πr²
V’ = 6πr (2 – r)
r = 0, r = 2
Sol. 3
r=2
V’ = 12πr – 6πr²
V’’ = 12π – 12πr
V’’ < 0
MAX
r
h
6
h = 6 – 2r
h = 6 – 2(2)
h=2
V = π(2)²(2) = 8π units³
3
1972 AB 4 BC 3
A man has 340 yards of fencing for enclosing two
separate fields, one of which is to be a rectangle
twice as long as it is wide and the other a square.
The square field must contain at least 100 square
yards and the rectangular one must contain at
least 800 square yards.
a. If x is the width of the rectangular field, what
are the maximum and minimum possible
values of x?
b. What is the greatest number of square yards
that can be enclosed in the two fields? Justify
your answer.
1972 AB 4 BC 3
2x
side length of square =
(340 – 6x) / 4
x
1972 AB 4 BC 3
a. Total Fence = 340 yards
2x² ≥ 800
x² ≥ 400
x ≥ 20 yards
[(340 – 6x) / 4]² ≥ 100
(340 – 6x) / 4 ≥ 10
340 – 6x ≥ 40
-6x ≥ -300
x ≤ 50 yards
2x
side length of
square =
(340 – 6x) / 4
x
20 ≤ x ≤ 50
1972 AB 4 BC 3
b. Let A be the total area of the rectangle and the square
A = 2x² + [(340 – 6x) / 4]²
A’ = (17/2)x – 255
0 = (17/2)x – 255
x = 30
2x
side length of
square =
(340 – 6x) / 4
x
A(20) = 3,825
A(30) = 3,400
A(50) = 5,100
Maximum Area = 5,100 yards²
when x = 50
Works Cited
• http://etc.usf.edu/clipart/41600/41699/fc_co
ne_41699.htm
• http://www.qcalculus.com/cal08.htm
• Calculus, James Stewart 5e
• http://office.microsoft.com/en-us/images/
© Kelsey Byrd, 2011