The sum of two nonnegative numbers is 36. Find the numbers if the difference of their square roots is a maximum.
$x + y = 36$ and
$S = \sqrt{x} - \sqrt{y}$
I keep coming up with X = 18 which gives me 0 in the 2nd derivative test but it is not a solution when substituted back into the 1st derivative. Is x = 18 therefore not a solution and there is no solution?
You want to maximise $|a-b|$ subject to $a^2+b^2=36$, knowing that $a$ and $b$ are non-negative. (I've changed the notation to avoid having lots of square roots)
Now $(a-b)^2=a^2-2ab+b^2=36-2ab$ with $ab\ge0$. It is clear that the maximum value of the right-hand side is $36$.
Let $x\ge y$. You need to find $x$ such that $$\sqrt x-\sqrt {36-x}$$is maximum. To do that by differentiating we obtain $$\dfrac{1}{2\sqrt x}-\dfrac{1}{2\sqrt {36-x}}=0$$which gives us $x=18$. But still two more candidates can be found on the boundaries where $(x,y)=(36,0)$ or $(x,y)=(0,36)$ therefore $$x=36\\y=0$$is the correct answer where $S=6$ at its maximal value.
We can and do assume $0\le x\le y\le 36$. Then $$ S(x,y)=\sqrt y-\sqrt x\le \sqrt {36} -\sqrt 0=6-0=6\ , $$ and this is the maximal value of the expression $S(x,y)$, it is obtained if and only if we have equality in the $\le$, i.e. $y=36$, $x=0$.
Note: We can better take $x=y=18$, this is a simpler situation, because the expression is not maximal, and in this case we do not need to find anything. (The question was "Find the numbers if the difference...") The problem should give the information if the difference of the square roots of $x,y$ is indeed maximal, so that we can decide if we want to find $x,y$. We may try to find the values in both cases, but in case we do not have maximality, we also do not have further information.
Given $0\le x,y\le 36$, denote: $x=a^2, y=b^2, 0\le a,b\le 6$, then: $$a^2+b^2=36 \ \ \text{(circle with center $(0,0)$ and radius $6$)}\\ z=a-b \ \ \text{to be maximized}$$ Refer to the graph:
$\hspace{3cm}$
The contour line is $b=a-z$ and $z$ achieves its minimum $-6$ at the corner (border) point $A(0,6)$ and maximum $6$ at the point $B(6,0)$.
In terms of $x,y$ they are: $$z(0,36)=-6 \ \text{(min)}; \ \ z(36,0)=6 \ \text{(max)}.$$
