Calculus - How to minimize x^2 + (1/x^2)

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In summary, the task was to find a positive number that would result in the sum of its square and its reciprocal being a minimum. The solution involved finding the derivative of the function y=x^2+1/x^2, setting it equal to 0, and solving for the critical point. The answer obtained was x=1, which is the only possible solution since the function is strictly convex in the positive region. Therefore, the minimum value is 1.
  • #1
yoyo16
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Homework Statement



Find a positive number such that the sum of the square of the number and its reciprocal is a minimum.

Homework Equations





The Attempt at a Solution


y=X^2+(1/X^2)
Derivative=2x-2x^(-3)
0=2x-2x^3
2/x^3=2x
2=2x^4
1=x^4
x=1

The answer I got was 1 but I'm not sure if this is right. Can someone please tell me if I did it right or wrong. If wrong, can you point out my mistake. Thanks :)
 
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  • #2
[strike]Looks good except that you missed a possible solution. What other value of x satisfies x^4=1?[/strike]

Oops, never mind. I didn't notice you were looking for a positive number.

You just now need to verify that the critical point corresponds to a minimum.
 
  • #3
Is it possible for the minimum to be a smaller value or would the smallest value be 1?
 
  • #4
yoyo16 said:
Is it possible for the minimum to be a smaller value or would the smallest value be 1?

If the graph is concave up for ##x>0## your relative minimum will be an absolute minimum on that interval. Have you checked that?
 
  • #5
yoyo16 said:

Homework Statement



Find a positive number such that the sum of the square of the number and its reciprocal is a minimum.

Homework Equations





The Attempt at a Solution


y=X^2+(1/X^2)
Derivative=2x-2x^(-3)
0=2x-2x^3
2/x^3=2x
2=2x^4
1=x^4
x=1

The answer I got was 1 but I'm not sure if this is right. Can someone please tell me if I did it right or wrong. If wrong, can you point out my mistake. Thanks :)

Your function is strictly convex in the convex region {x > 0}, so any stationary point is automatically a global minimum within the region.
 
  • #6
Oh ok so since it is absolute it would be 1 right?
 

FAQ: Calculus - How to minimize x^2 + (1/x^2)

1. What is the purpose of minimizing x^2 + (1/x^2) in calculus?

The purpose of minimizing x^2 + (1/x^2) in calculus is to find the minimum value of the function. This can help us determine the optimal value of x that will produce the lowest possible output for the function.

2. How do you find the minimum value of x^2 + (1/x^2) using calculus?

To find the minimum value of x^2 + (1/x^2) using calculus, we can take the derivative of the function and set it equal to zero. Then, we can solve for x to find the critical points. We can then use the second derivative test to determine if the critical point is a minimum or maximum.

3. What is the significance of the critical points in minimizing x^2 + (1/x^2)?

The critical points in minimizing x^2 + (1/x^2) are important because they represent the points where the derivative of the function is equal to zero. This means that the slope of the function at these points is flat, and they can help us determine the minimum value of the function.

4. Can calculus be used to minimize x^2 + (1/x^2) in real-life scenarios?

Yes, calculus can be used to minimize x^2 + (1/x^2) in real-life scenarios. For example, this function can be used to minimize the cost of materials for a given project or to optimize the production of a certain product.

5. Are there any alternative methods to minimizing x^2 + (1/x^2) without using calculus?

There are some alternative methods to minimizing x^2 + (1/x^2) without using calculus, such as graphing the function and finding the lowest point on the graph or using trial and error to find the minimum value. However, these methods may not be as efficient or accurate as using calculus.

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