Sneaky Square Roots

All expressions under the radicals are non-negative. If $x>y$ , then the last term shows $y\geq 140$ , and the first term shows $x\leq 20$ . However, this implies $140\leq y<x\leq 20$ , which is absurd.

If $x<y$ , then $y\leq140$ and $x\geq20$ . If none of the equality signs is achieved, the term under the middle radical is negative. If $y=140$ , then the expression reduces to $\sqrt{(x-20)(140-x)}$ which reaches its maximal value of 60 when $x=80$ since the term inside the radical is a quadratic function with coefficient quadratic term less than 0. If $x=20$ , then similarly the maximal value of 60 is reached when $y=80$

If $x=y$ , then the expression becomes $\sqrt{(140-x)(20-x)}$ The term inside is a quadratics opening upwards, so the maximal is reached at end points of the interval definition, which is the intersection of the intervals $[-40,100]$ and $[-20,200]$ . At -20, the function achieves the value of 80. At 100, the function has value less than 80, so the maximal is 80

Let $a=x-20, b=y-x, c=140-y$ . Let $S=\sqrt{(x-20)(y-x)}+\sqrt{(140-y)(20-x)}+\sqrt{(x-y)(y-140)}=\sqrt{ab}+\sqrt{-ac}+\sqrt{bc}$ . We claim that at least one of $a,b,c=0$ , otherwise, $S$ will not be a real number. When $a=0, x=20, S=\sqrt{(20-y)(y-140)}=\sqrt{-(y-80)^2+3600}, S_{max}=60$ . When $c=0$ , similarly we obtain $S_{max}=60$ . When $b=0, x=y$ , $S=\sqrt{(140-x)(20-x)}=\sqrt{(x-80)^2-3600}$ , $S$ achieve its maximum value when $|x-80|$ is maximum, since $x=y$ , $-20\leq x\leq100$ , $|x-80|_{max}=100$ , $S_{max}=80$ . Combining all three cases, $S_{max}=80$ .

Let us write $A = x-20$ , $B = y-x$ and $C = y-140$ . We first establish that one of these expressions must equal zero.

Observe that each of the terms in the given expression must either be real, or have a positive imaginary part (if the product within is negative). Thus, if any term has an imaginary part the entire expression will also have an imaginary part. Hence, we need each term to individually be real - and thus the product within the root must be positive. The given expression may be re-written as $\sqrt{AB} + \sqrt{AC} + \sqrt{-BC}$ Suppose all three terms are non-zero. Then, we have two cases - either $A$ is positive or negative. If $A$ is positive, then $B$ and $C$ must also be positive so that $\sqrt{AB}$ and $\sqrt{BC}$ are real. However, then $-BC$ will be negative and $\sqrt{-BC}$ is imaginary - a contradiction. Similarly if $A$ is negative then $B$ and $C$ must also be negative, and then $-BC$ will be negative and $\sqrt{-BC}$ is imaginary. Thus, at least one of $A$ , $B$ or $C$ must equal zero.

We now use casework on the three possibilities.

Case 1 : $A = 0$ . Then $x = 20$ . Our expression readily simplifies to $\sqrt{(20 - y)(y - 140)}$ . This expands to give us $\sqrt{-y^2 + 160y - 2800}$ , and through differentiation (for example) we find that the maximum value of $60$ occurs when $y = 80$ .

Case 2 : $B = 0$ . Then $x = y$ . Our expression simplifies to $\sqrt{(20 - x)(140 - x)}$ . This expands to yield $\sqrt{2800 + x^2 - 160x}$ . This quadratic is concave up with a minimum point at $x = 80$ (can find via differentiation). The possible values of $x$ here would be the intersection of the constraints over $x$ and $y$ . So $-20 \leq x \leq 100$ here. To get the maximum value, it occurs at the more distant endpoint in terms of x-coordinates, or at $x = -20$ . We find our maximum value for this case as $\sqrt{6400} = 80$ by substitution.

Case 3 : $C = 0$ . So $y = 140$ . Our expression reduces to $\sqrt{(x - 20)(140 - x)}$ and this is essentially the same quadratic as in Case 1 except with a change of variables - the maximum value of $60$ is unaffected since $x = 80$ is still in the allowable range.

Thus considering all of the three cases, the overall maximum value of the given expression is $\boxed{80}$ .

Let $A = \sqrt{(x-20)(y-x)}+\sqrt{(140-y)(20-x)}+\sqrt{(x-y)(y-140)}$

i) If $x=y$ , then we want to maximize $\sqrt{(140-x)(20-x)} = \sqrt{(x-80)^2-3600}$ where $-20 \leq x \leq100$ (the overlapped interval between $x,y$ ). Clearly, it is maximized on the furthest point from $80$ which is $x = -20$ . This yields $A = \sqrt{100^2-3600} = 80$

ii) if $x \neq y$ , for the first root we must have either a] $x = 20$ and $A = \sqrt{(20-y)(y-140)} = \sqrt{-(y-80)^2+3600}$ . To maximize $A$ , we need to minimize $(y-80)^2$ , so $y = 80 \Rightarrow A = 60$

b] $y = 140$ and $A = \sqrt{(x-20)(140-x)} \sqrt{-(x-80)^2+3600}$ . By similar reasoning as the previous case, $x = 80 \Rightarrow A = 60$

c] $x-20 > 0, y-x > 0 \Rightarrow x > 20 , y > x$ and from the 3rd root we get $y < 140$ . But then $140-y > 0, 20-x < 0$ and this will imply that the term inside the 2nd root is negative, contradiction.

d] $x-20 < 0, y-x < 0 \Rightarrow x < 20, y < x$ and from the 3rd root, we get $y > 140$ . But this means $20 > x > y > 140$ , a contradiction

Thus, the maximal value is $80$

let a=x-20 , b=y-x ,c=160-y;
so ab>=0,-ac>=0,bc>=0;
so a=0 or b=0 or c=0;
so max =80

Seeing all of the square roots, it seems that in each square root, each multiple inside it has to be either both positive or both negative. Since x- 20 and 20-x both have to be either positive positive or negative negative or negative positive or positive negative, the only value of x that maximizes both square roots is x = 20. Now we are left with sqrt((20-y)(y-140)), and the max value of this is y = 80, getting a value of 60. Now let's try the same thing with 140-y and y-140 and see if we get a larger answer. Doing the same steps above, we get y = 140, and x = 80, and again we achieve an answer of 60. Now we try one last step, let's try making x = y. With that we achieve the sqrt((140-y)(20-x)) = sqrt((140-x)(20-x)). The domain of this is x is less than or equal to 20 or greater than or equal to 140. so seeing our restrictions, x can only be a maximum of 100 and a minimum of -40. However since x = y, the only values of x we can use are from -20 to 100. Inputting -20 gets us a value of 80, and inputting 100 doesn't get us an integer (remember our answer must be an integer from 0 to 999!). Thus, the max value is 80.

If x>y, to make the 1st term a real no. x should be less than -20 as y-x will be a negative value. but this case contradicts because of the third term as x-y will be positive but y-140 will be negative because y is negative in this case. if y>x , the first term will be real only if x>20 and to make the 2nd term a real no. we have to y should be greater than 140. Due to this y-140 in term 3 becomes positive and alongside with this x-y becomes negative so this too contradicts. the last and only case remaining is when x = y. in this case the 1st and the 3rd terms become 0 only the 2nd term remains. to maximize the 2nd term we have put the values of x and y as -20(we cannot put it as -40 because the the minimum value of y is -20). after that when we solve the expression by putting the values of x and y as -20 we get 80 as the result.

First, we should observe that we could rewrite the expression as follows: $\sqrt{-(x-20)(x-y)}+\sqrt{(y-140)(x-20)}+\sqrt{(x-y)(y-140)}$ . Also, all the products must be grater or equal to 0. If x and y aren't equal, there is no way to achive that, because if $x-y>0$ , $y-140$ must be positive (from the third root). From the second root, we see that $x-20$ must be positive too. But this means that $-(x-20)(x-y)$ is smaller than 0. Simillarly, $x-y$ can't be smaller than 0. So, $x-y=0$ which means that $x=y$ . If $x=y$ , the relation becomes equal to $\sqrt{(140-x)(20-x)}=\sqrt{x^2-160x+2800}$ , $-20<=x<=100$ . Now all we have to do is to find the maximum value of $\sqrt{x^2-160x+2800}$ . It's maximum value is equal to $\sqrt{a}$ , where $a=max(x^2-160x+2800)$ , $-20<=x<=100$ and a must be positive. That is a parabola, so it's maximum value is ethier at -20, ethier at 100 on that interval. If $x=100$ , $x^2-160x+2800$ is equal to -3200, so we try $x=-20$ , which gave us the value 6400. $\sqrt{6400}=80.$

1) Suppose x > y:
In the third part, $\sqrt{(x-y)(y-140)}$ , "(x - y)" is positive, so "(y - 140)" must be positive too.
Consequence: y > 140

2)Take the middle part $\sqrt{((20 - x)(140 - y))}$
=> "(140 - y)" is negative (due to the consequence above), so "(20 - x)" must be negative too.
Consequence: x > 20

3) Take the first part $\sqrt{(x - 20)(y - x))}$
The first term is positive, since x > 20
The second term is negative, since x > y
That's nonsensical, due to the square root.

• (...) - Analogous case to y > x

  That way, x must be equal to y.

• Calculating the equation, we get a 0 in the first and third piece, and it rests a $\sqrt{(20 - x)(140 - x)}$ , which is maximum when "x" and "y" are minimum.
• The lowest "x" is -40
• The lowest "y" is -20
  Thus, once x = y, the lowest number they may be is -20.
  Calculating it, we get a 80.
  

Let $a = x - 20, b = y - 140, c = x - y$ . Then the expression in question becomes (\ \sqrt{-ac} + \sqrt{ab} + \sqrt{bc} ). Doing some case analysis on the variables reveals that they cannot be purely positive or negative, so at least one of them has to be $0$ , but we're looking for the maximum value of the expression, so we don't want zeroes across the board. In that case, we know that one of them should be $0$ . We also see that the original middle root has some large numbers in it, so we leave that one as the nonzero one. To achieve this, we must have that $c = 0$ , or $x = y$ . Now we're left to maximize $\sqrt{(140 - y)(20 - y)}$ , where $-20 \leq y \leq 100$ . A bit of experimentation at this point yields the optimal value, $y = x = -20$ .

Let x-20=a, y-x=b, 140-y=c. Then√(x−20)(y−x)+√(140-y)(20-x)+√(x-y)(y-140)=√ab+√-ac+√bc . ab,-ac,bc must all be none negative number

By looking at -ac and bc ,we know that a and b can't be both positive or both negative if you want them to be none negative number. By looking at ab, we know that a and b can't be one positive one negative One of a and b must be zero.

If a=0,x=20,√ab+√-ac+√bc=√bc =√(140-y)(y-20)

max value:when b=c (80)

If b=0,x=y,√ab+√-ac+√bc=√-ac =√(140-y)(20-x)

max value:when x and y are smallest, because x=y, therefore x=y=-20 (80)

If a=b=0,then√ab+√-ac+√bc=0

Largest possible real value for this question is 80

When x=20, S = $\sqrt{(20-y)(y-140)}$ , here max value can be $\sqrt{(-60)(-60)}$ = 60 when y = 80

When y=140, S = $\sqrt{(x-20)(140-x)}$ , here max value can be $\sqrt{(60)(60)}$ = 60 when x = 80

When x=y, S = $\sqrt{(140-y)(20-y)}$ , here max value can be $\sqrt{(160)(40)}$ = 80 when y = -20 since y can't > 100 as x= y.

So ans is 80.

We can not have x < y or y < x, only * x = y * (positive argument for sqrt). In this case, the expression remain sqrt [(140 - x) (20 - x)], with -20 <= x <= 20. sqrt [(120+ 20-x ) (20 -x)] is maximum when 20 - x is maximum, ie for x = -20. The requested value will be 80 :)