Let


\(f(a)=\sqrt{a}+\sqrt{170-a}\)


with \(75<a<85\) .
Notice that


\(f(80)=\sqrt{80}+\sqrt{90}\)


so Quantity B is exactly f(80) .
Now check the behavior of f(a) on \(75<a<85\) . Its derivative is


\(f^{\prime}(a)=\frac{1}{2 \sqrt{a}}-\frac{1}{2 \sqrt{170-a}}\) .


Since \(a<85\) , we have \(a<170-a\) , so \(\sqrt{a}<\sqrt{170-a}\) , hence


\(\frac{1}{\sqrt{a}}>\frac{1}{\sqrt{170-a}}\),


which means \( f^{\prime}(a)>0\) . So f is strictly increasing on this interval.
Therefore:

If a<80 , then \(f(a)<f(80)\) , so Quantity A < Quantity B.
If a=80 , then \(f(a)=f(80)\) , so Quantity \( \mathrm{A}=\) Quantity B.
If a>80 , then \( f(a)>f(80)\) , so Quantity \( \mathrm{A}>\) Quantity B.


Since $a$ can be any number in (75,85) , including values on both sides of 80 , the comparison is not fixed.

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