• Support PF! Buy your school textbooks, materials and every day products Here!

Extreme values problem

  • Thread starter Nitrate
  • Start date
  • #1
75
0

Homework Statement


find two positive numbers with product of 200 such that the sum of one number and twice the second number is as small as possible.



2. The attempt at a solution

my work:
xy=200 ==> y = 200/x
x+2y = s (what we need to minimize)
x+2(200/x) =s
x+400x^-1 = s
1-400x^-2 = ds/dx
(x^2-400)/x^2 = dx/dx
(x-200)(x+200)/(x^2) = ds/dx
crit numbers: 0, 200, -200 (not included because the domain is x>0)
 

Answers and Replies

  • #2
22,097
3,279

Homework Statement


find two positive numbers with product of 200 such that the sum of one number and twice the second number is as small as possible.



2. The attempt at a solution

my work:
xy=200 ==> y = 200/x
x+2y = s (what we need to minimize)
x+2(200/x) =s
x+400x^-1 = s
1-400x^-2 = ds/dx
(x^2-400)/x^2 = dx/dx
(x-200)(x+200)/(x^2) = ds/dx
crit numbers: 0, 200, -200 (not included because the domain is x>0)
Are you sure that [itex]x^2-400=(x-200)(x+200)[/itex]. What is 200*200??
 
  • #3
75
0
Are you sure that [itex]x^2-400=(x-200)(x+200)[/itex]. What is 200*200??
oh wow >.> what a silly error.
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


find two positive numbers with product of 200 such that the sum of one number and twice the second number is as small as possible.



2. The attempt at a solution

my work:
xy=200 ==> y = 200/x
x+2y = s (what we need to minimize)
x+2(200/x) =s
x+400x^-1 = s
1-400x^-2 = ds/dx
(x^2-400)/x^2 = dx/dx
(x-200)(x+200)/(x^2) = ds/dx
crit numbers: 0, 200, -200 (not included because the domain is x>0)
Here is a little hint that applies to ANY problem of the form min f(x) = Ax + B/x with A,B>0 (and we want x > 0). At the min, both terms of f are *equal*, so Ax = B/x. That means that x = sqrt(B/A). (Remembering equality of the two terms is easier than remembering the final formula.)

By the way, that "equality" result follows from calculus, but can also be obtained without using calculus---that is the basis of so-called "Geometric Programming".

RGV
 

Related Threads on Extreme values problem

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
13K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
16
Views
6K
  • Last Post
Replies
2
Views
777
Top