MATH241 Lecture 27: Optimization Problems
9 May 2018
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MATH241 - Lecture 27 - Optimization Problems
4.7: Optimization Problems
Steps in Solving Optimization Problems
1) Understand the problem
2) Draw a diagram
3) Introduce notation
4) Express the quantity to be optimized in terms of the variables
5) Use the given information to rewrite the quantity to be optimized as a function of a
single variable. Determine its domain
6) Use the methods of Sections 4.1 - 4.3 to find the absolute extremum. If the domain is
closed interval use the closed interval method
Example
:
For two non-negative numbers, twice the first plus the second is 12. What is the maximum
product of the two numbers?
Solution:
Let and represent two non-negative numbers, and let x y yP =x
x22 + y= 1
→ 2xy = 1 − 2
Substitute 2xP =x(12 x)− 2 = 1 − 2x2
If and then the domain of is x≥ 0 y≥ 0 P(x) 0, 6
[ ]
The domain is closed so use the closed interval method. Find the critical numbers
2xP ′(x) = 1 − 4 = 0
→ → x24 = 1 x= 3
(0) 2P= 0 · 1 = 0
8P(3) = 3 · 6 = 1
P(6) = 6 · 0 = 0
The maximum product of the two numbers is 18 when and x= 3 y= 6
Example
:
A fence is to be built to enclose a rectangular area of . Three sides of the fence will be00 ft82
made using material costing $6 per linear foot and the fourth side will use material costing $18
per linear foot. Find the most economical dimensions
Solution:
Area of a rectangle: y00A=x= 8
Without loss of generality (WLoG) let the
bottom cost $18/foot and the other three sides
cost $6/foot
The the cost is x y 8x yC = 6 + 6 + 1 + 6
4x2y= 2 + 1
Express as a function of one variable → 00 y8 = x y =x
800
Substitute 4x2 C(x) = 2 + 1 (x
800 )
4x= 2 + x
9600
The domain is (0, ∞)
Find the critical numbers:
4C′(x) = 2 − x2
9600 = 0
→ 4
x2
9600 = 2
→ 60024x2= 9
→ x2=24
9600
→ 00x2= 4
The critical number is 0x= 2
If then 0x= 2 0y=20
800 = 4
Can’t use the closed interval method so we must use the first derivative test of the second
derivative test
Document Summary
Steps in solving optimization problems: understand the problem, draw a diagram. Introduce notation: express the quantity to be optimized in terms of the variables, use the given information to rewrite the quantity to be optimized as a function of a single variable. Determine its domain: use the methods of sections 4. 1 - 4. 3 to find the absolute extremum. If the domain is closed interval use the closed interval method. For two non-negative numbers, twice the first plus the second is 12. Let x represent two non-negative numbers, and let y and x. If and y 0 then the domain of. The domain is closed so use the closed interval method. P (x) = 1 4 = 0 x x = 3 x. P (3) = 3 6 = 1. P (6) = 6 0 = 0. The maximum product of the two numbers is 18 when x = 3 and y = 6.
