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

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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)
 
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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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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