The sum of two positive numbers is 20. Find the numbers if the sum of their squares is as large as possible; as small as possible.
What does it mean? This is from applications of derivatives. How do I solve this.
Two numbers, x, and 20-x
Sumsquares= x^2 + (20-x)^2 is max.
Take the derivative of this, set to zero, solve for x, and 20-x. You should get two solutions, one is a max, one is a min.
I get x=10, means the numbers are 10 and 10.
At the back of the book , answers are:
as large as possible 0,20
as small as possible 10,10
How do we get the second solution? that is 0 and 20
Jen, I have a problem with the answer. The problem was stated as two positive numbers. Zero is not a positive number, positive numbers are greater than zero.
Now, your question. You can not use the derivative except to find where the slope is zero (max or min). Therefore, the other solution has to be tested at at the endpoints in the original function. The slope is not zero at zero, or x=20.
we have a similar problem, except the questions asks for "nonnegative numbers" which will get you the answers that are at the back of the book.
perhaps just a miscommunications on the book's or Jen's part?
It seems that there might be an error in the given answers or a miscommunication. The problem states that the two numbers are positive, so zero would not be considered a valid solution.
To find the numbers if the sum of their squares is as large as possible, you can indeed use derivatives. Let's call the two numbers x and 20-x.
The sum of their squares is given by the function f(x) = x^2 + (20-x)^2. To find the maximum value of this function, you can take its derivative and set it equal to zero:
f'(x) = 2x - 2(20 - x) = 0
Simplifying, we get:
2x - 40 + 2x = 0
4x = 40
x = 10
So, one of the numbers is 10 and the other number is 20-10 = 10. This means that the sum of their squares is as large as possible when both numbers are 10.
However, as you correctly pointed out, this does not match the given answers from the book, which state that the numbers are 0 and 20. This seems to be contradictory since zero is not considered a positive number.
In this case, it's possible that there is an error in the book or a miscommunication. To clarify, you can reach out to the book's author or your instructor for further clarification on the answer key.
Yes, it seems there is a miscommunication between the book and Jen. The problem states that we are looking for two positive numbers, which means they must be greater than zero. However, the answer provided in the book includes the numbers 0 and 20, which contradicts the given condition.
To solve the problem correctly, we can follow the steps you mentioned:
1. Let the two positive numbers be x and 20 - x.
2. The sum of their squares is given by (x^2) + ((20 - x)^2).
3. To find the values of x that maximize or minimize this sum, we can take the derivative of the sum with respect to x and set it equal to zero.
However, in this case, we are only interested in knowing if there is another solution apart from x = 10 and 20 - x = 10. To check this, we can evaluate the sum of squares at the endpoints x = 0 and x = 20.
For x = 0: (0^2) + ((20 - 0)^2) = 400
For x = 20: (20^2) + ((20 - 20)^2) = 400
We can see that these two endpoints yield the same sum of squares value as the solution x = 10. Therefore, the other solution would be (0, 20), as indicated by the book.
However, it's worth mentioning that the book's answer contradicts the given condition of the numbers being positive. So, it's possible that there may be an error in the book or a misunderstanding in the interpretation of the problem.