I have no clue where to start.

The sum of two numbers is $28$. Find the numbers assuming that 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.


  • 3
    $\begingroup$ 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? $\endgroup$ – xbh Sep 25 '18 at 16:37
  • 2
    $\begingroup$ Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not? $\endgroup$ – amWhy Sep 25 '18 at 16:39
  • 1
    $\begingroup$ Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $ $\endgroup$ – amWhy Sep 25 '18 at 16:40
  • 1
    $\begingroup$ 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 $\endgroup$ – Zebert Sep 25 '18 at 16:42
  • $\begingroup$ Would factoring it help? $\endgroup$ – Zebert Sep 25 '18 at 16:43

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:

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

  • $\begingroup$ btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio. $\endgroup$ – Finn Eggers Sep 25 '18 at 16:46


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?"


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.