I have no clue where to start.

The sum of two numbers is 28. Find the numbers if the sum of their squares is a minimum

I am an eleventh-grader. I only learned how to find the minimum for functions like $y=ax^2+bx+c$ or $y=a(x-h)^2+K.$ I don't believe I know how to do calculus and don't know what a derivative is.

Thanks!

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    We have $x+y =28$. We are requested to find $x,y$ s.t. $x^2+y^2$ attains minimum. Now $y = 28-x$, so…… could you see how to proceed now? – xbh Sep 25 at 16:37
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    Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not? – amWhy Sep 25 at 16:39
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    Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $ – amWhy Sep 25 at 16:40
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    No, can you continue? I only learned how to find min in y=ax^2+bx+c or y=a(x-h)^2+K. I don't believe I know how to do calculus and don't know what a dervative is – Zebert Sep 25 at 16:42
  • Would factoring it help? – Zebert Sep 25 at 16:43
up vote 3 down vote accepted

Let $x, y$ be such that $x+y = 28 \iff y= 28-x$.

Then, the function we want to minimize looks something like this: $$x^2 + y^2 = x^2 +(28-x)^2 = x^2+ 784-56 x + x^2\tag{1}$$

Simplify $(1)$ to get $$f(x) = 2x^2 -56x + 784\tag{2}$$ And given you know how to find the minimum of a function of the form $$f(x) = ax^2 +bx+c$$ use your knowledge to minimize the function $(2)$

Begin by writing down what the problem is asking you to do.

The sum of two numbers is 28

So we take two numbers, $x$ and $y$. From this, we know that $x + y = 28$.

the sum of their squares is a minimum

Here, we want to minimize $S = x^2 + y^2$. From the above, we can see that $y = 28 - x$, which by substituting this into $S$, we get $$\begin{align}S &= x^2 + (28 - x)^2 \\ &= 2x^2-56x + 784.\end{align}$$ This is a quadratic equation, which the minimum or maximum of the parabola (for any $y = ax^2 + bx + c$) occurs at $x = -{b\over 2a}$. Here, $a = 2$ and $b = -56$. Hence, $$\begin{align}x &= -{-56\over 2(2)} \\ &= 14.\end{align}$$

Here we have found $x$. I will leave it to you to find $y$.

By the RMS-AM inequality (root-mean square vs. arithmetic mean):

$$\sqrt{\frac{a^2+b^2}{2}} \ge \frac{|a|+|b|}{2} \ge \frac{a+b}{2} = \frac{28}{2}=14$$

Equalities hold iff $\,a=b \ge 0\,$, therefore the minimum of $\,a^2+b^2\,$ is attained for $\,a=b=14\,$.

Let's name the numbers $a$ and $b$.

$$a + b = 28$$

Furthermore, let's define the sum of the squares (so we can minimize that):

$$f(a,b) = a^2 + b^2$$

Let's use the first formula to reduce the amount of variables:

$$f(a)=a^2 + (28 - a)^2 = 2 a^2 - 2 \cdot28\cdot a +28^2$$

now let's find the minimum of $f(x)$:

$$\dfrac{df(a)}{da} = 4a-2\cdot28$$

Finally, we will set this to zero and solve for a:

$$4a-2\cdot28=0$$ $$4a=2\cdot28$$ $$a = 14$$

  • btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio. – Finn Eggers Sep 25 at 16:46

$$\min_{x+y=28}x^2+y^2$$

is the same as

$$\min_xx^2+(28-x)^2$$ which you find by differentiation.

$$x-(28-x)=0\implies x=14.$$

Note that $$(a-b)^2 \ge 0 \implies a^2+b^2 \ge 2ab \implies 2(a^2+b^2) \ge a^2+b^2 +2ab.$$ That is, $$a^2+b^2 \ge\frac{(a+b)^2}{2},$$ where the equality holds when $a-b=0$, or equivalently $a = b$. In your case, $a+b=28$. Therefore, $a^2+b^2$ will be minimum when $a = b = 28/2 = 14$.

"I have no clue where to start" "I don't believe I know how to do calculus and don't know what a derivative is. "

Then experiment.

$a + b =28$. So you could have $14 + 14 = 28$ and $14^2 + 14^2 = 196 + 196 = 392$. Or you could have $1 + 27 = 28$ and $1^2 + 27^2 = 1 + 729 = 730 > 392$. Or you can have $13 + 15 = 28$ and $13^2 + 15^2 = 169 +225 = 394 > 392$.

Can you make a hypothesis? Can you argue why.

Hint: Have you ever hear of the A.M/G.M inequality? (If $a> 0; b> 0$ then $\sqrt ab \le \frac {a+b}2$ Can you see how that would be relevent.)

I only learned how to find the minimum for functions like y=ax2+bx+c or y=a(x−h)2+K

Well, that's more than I knew in the 11th grade!

Consider $a + b = 28$ so $b = 28 -a$ and you want to find the minimum of $a^2 + b^2 = a^2 + (28 -a)^2= 28^3 - 2*28a + 2a^2$. Can you use what you know to find the minimum?

but... experiment and muck around and see what you find. No one is expectiong you to get it right away.

====

If it were up to me and I only had the knowledge I had in the $11$th grade I would notice that if $a$ and $b$ are close together it seems that the sums of the squares are smaller than than if the are far apart. I would then figure if $a = b = 14$ would be the smallest $a^2 + b^2$.

Then I'd try to see if I can justify it. I'd figure $a + b = 28$ so $b = 28-a$ so I am trying to minimalize $a^2 + b^2 = a^2 + (28 - a)^2 = 2a^2 - 56a + 28^2$. That is smallest when $2a^2 - 56a$ or when $a^2 - 28a = a(a-28)$ is smallest.

If $a$ goes up by, say $1$ then $a'^2 - 28a' = (a + 1)^2 - 28(a+1) = a^2 - 28 + (2a + 1)-28 $. That's an increase if $(2a + 1) -28$ is positive and a decrease if $(2a+1) - 28$ is negative.

In other words if $2a + 1 -28 > 0$ or $2a > 27$ or $a > 13.5$ then be increasing by $a$ by one will increase $a^2 + b^2$ but it if $a < 13.5$ then increasing $a$ by one will be decreasing the value of $a^2 + b^2$. So if $a = 13.5$ is when adding $1$ to $a$ will stop decreasing $a^2 + b^2$ and start increasing it. So it's reasonable that somewhere near $a = 13.5$ and $b = 14.5$ that $a^2 + b^2$ reaches a minimum and has nowhere to go but up.

I suppose I'd then think that $1$ was a jerky to big of a jump and I'd consider a jump of $d$ where $d$ can be something small like $\frac 1{1000}$. Then jumping from $a$ to $a + d$ would change $a^2 -28a$ to $(a + d)^2 - 28(a+d) = a^2 - 28a + 2ad + d^2 - 28d$ or a differencd of $2ad + d^2 - 28d$. So to find the point where that "stops" being a negative decrease and starts being a poisitive increase I want to find when $2ad + d^2 - 28d > 0$ and when $2ad + d^2 - 28d =0$ and when $2ad + d^2 - 38d < 0$.

If we divide by $d$ (which is a tiny positive amount) in other words when $2a + d > 28$ and $2a + d = 28$ and if $2a + d < 28$. Or with $a > 14 - \frac d 2$ or $a = 14 - \frac d2$ or $a , 14 - \frac d2$. As $d$ can be very small this shift when we go from this being a decrease to this being an increase is at $a = 14$

Now maybe it would have occured to me (I honestly don't know if it would have or not) that if we can replace $b$ with $28 -a$ we could "center" by averaging $\frac {a+b}2 = \frac {28}2 = 14$ and replacing $a = 14 -e$ and $b = 14 + e$ so $a + b = (14 -e) + (14 + e) = 28$.

So now we want to minimize $(14 - e)^2 + (14 + e)^2 = 14^2 - 28 e + e^2 + 14^2 + 2e + e^2 = 2*14^2 + 2e^2$.

How do we minimize that? Well $e^2 \ge 0$ so to minimize it we simply take $e = 0$. So it is minimum at $e =0$ or $a = 14 - 0 = 14 = b = 14 + 0$.

Suppose you take a line segment of length 28, and divide it into two pieces, $left$ and $right$, with the $left$ part smaller than the $right$ part. Now draw the squares with sides $left$ and $right$. Now suppose you move the dividing point over by $h$ to the right. The $left$ square will increase by height $h$, while the $right$ square will lose $h$ in height. While the amount of height that $left$ gains is equal to the height that $right$ loses, $right$ loses that height over a greater distance, so the area that $right$ loses is more than the area $left$ gains.

The above hold until $left$ is equal to $right$. So if you want to decrease the total area, you should increase the smaller length until the lengths are equal. Hence, you want $left=right$, which happens when they're both equal to $14$.

Another way of looking at it is that if you have a right triangle, the length squared of the hypotenuse is the sum of the squares of the legs. So this is equivalent to "You have a right triangle whose legs add up to 28. What leg lengths minimize the hypotenuse?"

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