How Do You Minimize the Sum of One Number and Twice Another Given Their Product?

  • Thread starter Thread starter Nitrate
  • Start date Start date
Click For Summary
To minimize the sum of one number and twice another given their product of 200, the relationship xy=200 leads to y=200/x. The expression to minimize becomes s=x+400/x. The derivative ds/dx is calculated, revealing critical points at 0 and 200, with only 200 being relevant due to the domain restriction of positive numbers. A hint suggests that for minimization problems of the form Ax + B/x, the minimum occurs when Ax equals B/x, leading to the conclusion that x should equal the square root of B/A. This approach can be applied broadly in similar optimization problems.
Nitrate
Messages
75
Reaction score
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)
 
Physics news on Phys.org
Nitrate said:

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 x^2-400=(x-200)(x+200). What is 200*200??
 
micromass said:
Are you sure that x^2-400=(x-200)(x+200). What is 200*200??

oh wow >.> what a silly error.
 
Nitrate said:

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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
2
Views
1K
Replies
12
Views
4K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
2
Views
1K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K