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

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  • #1
Nitrate
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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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  • #2
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 [itex]x^2-400=(x-200)(x+200)[/itex]. What is 200*200??
 
  • #3
micromass said:
Are you sure that [itex]x^2-400=(x-200)(x+200)[/itex]. What is 200*200??

oh wow >.> what a silly error.
 
  • #4
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
 

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

1. What is an extreme values problem?

An extreme values problem, also known as an extreme value analysis or extreme value theory, is a statistical method used to analyze and predict the likelihood of rare and extreme events occurring in a given dataset.

2. What types of data are typically used in extreme values analysis?

Extreme values analysis is typically used with data that follows a continuous distribution, such as financial data, weather data, or environmental data. It can also be used with discrete data that has been transformed into a continuous distribution, such as counting data or binary data.

3. How is an extreme value problem different from a normal distribution problem?

An extreme value problem focuses on the tail end of a distribution, looking at the probability of rare events occurring. In contrast, a normal distribution problem looks at the overall shape and central tendency of a distribution.

4. What are some common applications of extreme values analysis?

Extreme values analysis is used in a variety of fields, including finance, insurance, climate science, and engineering. It can be used to assess the risk of extreme events, such as natural disasters or stock market crashes, and make informed decisions based on this risk.

5. What are the limitations of extreme values analysis?

Extreme values analysis assumes that the data follows a continuous distribution, which may not always be the case. It also relies on the assumption that extreme events are independent and identically distributed, which may not hold true in all situations. Additionally, extreme values analysis is limited by the amount and quality of data available, as it requires a sufficient amount of data to make accurate predictions.

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