# What is the least possible sum of squares when the sum of two numbers is 20?

• Incog
In summary, the least possible sum of the squares of two numbers whose sum is 20 is 200. This can be found by using the equation x + y = 20 and finding the minimum value of the expression x^2 + y^2, which is 200 when x = y = 10. This can also be proven by using the concept of completing the square.
Incog

## Homework Statement

The sum of two numbers is 20. What is the least possible sum of their squares.

2. The attempt at a solution

Before I show my work, I'm pretty sure I have the answer. I think it's 200. If you add 10 and 10, you will have 20. If you square 10 you get 100, thus the sum of the squares would be 200. If you used any other numbers to get a sum of 20 (i.e. 1 and 19, 2 and 18, 3 and 17, etc.), and you'd end up with a number over 200. For example, 18 + 2 = 20, 18^2 = 324, 2^2 = 4, 324 + 4 = 328. For all numbers other than 10 and 10, you'll get an answer over 200.

It's just I'm not exactly sure how to show my work for that. The only thing I can come up with is this:

The equation to show that the sum of two numbers equalling 20 is x + x = 20.

To show the squares of those numbers would be x^2 + x^2 = 20.

Therefore, 2x^2 = 20.

x^2 = 20/2

x^2 = 10

Am I on the right track with this?

I think so...just show that x is the square root of 10 then.

This problem can be done using derivatives...

let the numbers be say $$x$$ and $$a$$

Now $$x+a=20$$
also,

$$a=20-x$$

now we wish to find the minimum value of $$(20-x)^2+x^2$$

if we graph the equation...set any point on the y-axis = $$(20-x)^2+x^2$$

if we perform $$\frac{dy}{dx}$$ and set the slope to zero

we get

$$4x=40$$

so x=10 and therefore y=10

i guess...therefore the minimum sum of their squares is 200

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In general, you have an inequality:
$$\sqrt{\frac{x^2+y^2}{2}}\ge \frac{x+y}{2}$$

equality occurs if and only if x=y.
although this inequality only works for positive x and y. you can put absolute values around them for this problem.
$$\sqrt{\frac{x^2+y^2}{2}}\ge\frac{|x|+|y|}{2}\ge \frac{x+y}{2}$$

or root mean square >= arithmetic mean

as for proving it... just expand and simplify

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What a mess! There's a very simple solution.

If $$a + b = 20$$ then

$$a = 10 - c$$ and $$b = 10 + c$$. The sum of the squares is then

$$(10 - c)^2 + (10 + c)^2 = 200 + 2c^{2}$$.

Obviously we have to let c = 0 to get a minimum. Thus a = 10 and b = 10 and the the least possible sum is 200.

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Incog said:

## Homework Statement

The sum of two numbers is 20. What is the least possible sum of their squares.

2. The attempt at a solution

Before I show my work, I'm pretty sure I have the answer. I think it's 200. If you add 10 and 10, you will have 20. If you square 10 you get 100, thus the sum of the squares would be 200. If you used any other numbers to get a sum of 20 (i.e. 1 and 19, 2 and 18, 3 and 17, etc.), and you'd end up with a number over 200. For example, 18 + 2 = 20, 18^2 = 324, 2^2 = 4, 324 + 4 = 328. For all numbers other than 10 and 10, you'll get an answer over 200.

It's just I'm not exactly sure how to show my work for that. The only thing I can come up with is this:

The equation to show that the sum of two numbers equalling 20 is x + x = 20.
When you write this, you are assuming x= y which is the same as assuming x=y= 10. "The equation to show that the sum of two numbers equalling 20" is x+ y= 20.

To show the squares of those numbers would be x^2 + x^2 = 20.
No, nothing was said about the sum of the squares being 20! Even you are claiming it is 200!

[/quote]Therefore, 2x^2 = 20.

x^2 = 20/2

x^2 = 10

Am I on the right track with this?[/QUOTE]
Well, if x= y and x2= 10, then x2+ y2= 20 which isn't what you originally thought, is it? In fact, it comes from your incorrect equation 2x2= 20!

You know that x+ y= 20 so y= 20- x. Now, the sum of the squares is x2+ y[/sup]2[/sup]= x2+ (20- x)2= x2+ 400- 40x+ x2= 2x2- 40x+ 400.

Now, you can "complete" the square to find the maximum possible value of that.

Last edited by a moderator:
Alright I think I got it:

2x^2 - 40x + 400
= 2(x^2 - 20x) + 400
= 2[x^2 - 20x + (20/2)^2 - (20/2)^2] + 400
= 2[x^2 - 20x + 10^2 - 10^2] + 400
= 2[(x-10)^2 - 100] + 400
= 2(x-10)^2 + 400 - 200
= 2(x-10)^2 + 200

10 being the two squares
200 being the least possible sum

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First, it's x - 10 inside the bracket and second... this is a totally uselessly long way to do it.

Oh yeah, just saw that.

And that's the way the teacher taught it to us. What's the shorter way?

The way Werg22 showed you in post 5 is the best way.

Or you can find the axis of symmetry once you have y = 2x^2 - 40x + 400

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## 1. What is the difference between a sum and a square?

A sum is the result of adding two or more numbers together. A square is a number that is multiplied by itself, resulting in a product. For example, the sum of 3 and 5 is 8, while the square of 3 is 9.

## 2. Why is it important to find sums and squares?

Finding sums and squares is important in many mathematical and scientific applications. It allows us to solve equations, analyze data, and make predictions. Additionally, understanding sums and squares helps build a strong foundation for more complex mathematical concepts.

## 3. How do you calculate a sum?

To calculate a sum, simply add together all of the numbers given. For example, to find the sum of 2, 4, and 6, you would add 2 + 4 + 6 = 12.

## 4. How do you find the square of a number?

To find the square of a number, multiply the number by itself. For example, to find the square of 5, you would multiply 5 x 5 = 25.

## 5. What are some real-life applications of finding sums and squares?

Finding sums and squares is used in many fields, such as engineering, finance, and statistics. It is commonly used to analyze data, make predictions, and solve equations in various industries. For example, calculating the sum of sales data can help a business determine their profits for a given period, while finding the square of a number can be used to calculate the area of a room for construction purposes.

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