# Extreme values problem

## 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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## 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??

Are you sure that $x^2-400=(x-200)(x+200)$. What is 200*200??
oh wow >.> what a silly error.

Ray Vickson
Homework Helper
Dearly Missed

## 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