While tutoring my pre-cal tutees, I spontaneously created a problem about minimum value that involves a unit circle and a straight line $y = x – 2$. Specifically, I was asking my tutees the following question:
Question: Find the point on the unit circle center at the origin such that the distance from this point to the line $y = x – 2$ is minimum. Since this is a pre-calc course, calculus is not allowed.The question leads to the following problem:
Minimize $5 – 2ab – 4(a – b)$ subject to $a^2+b^2 = 1$ without using calculus.
For my part, I worked out the square distance $d^2$ formula from a point $(a,b)$ on the circle ($a^2 + b^2 = 1$) to the line $y = x – 2$, and I got the expression: $f(a,b) = 5 – 2ab – 4(a – b)$. I was trying to use AM-GM inequality but the second term and the third term seem to cancel out one another term effect. So AM-GM inequality seems not very helpful unless I miss something. Any suggestion or technique or inequality application that can be used here to tackle it?
Edit: For getting to $f(a,b)$, I used the distance formula from the point $P(a,b)$ to the line $y = x – 2$ or $x – y – 2 = 0$. If $d$ is the distance from the point $P$ above to the given line then $d = \dfrac{|a – b – 2|}{\sqrt{1^2+1^2}} = \dfrac{|a – b – 2|}{\sqrt{2}}\implies d^2 = \dfrac{(a – b – 2)^2}{2}= \dfrac{a^2+b^2+4 – 2ab -4a + 4b}{2}= \dfrac{5-2ab-4(a-b)}{2}$. Let $f(a,b) = 5 – 2ab – 4(a-b)$.
Best Answer
Let $a=\frac{1}{\sqrt2}$ and $b=-\frac{1}{\sqrt2}.$
Thus, we obtain a value $6-4\sqrt2.$
We'll prove that it's a minimal value.
Indeed, let $a=\frac{x}{\sqrt2}$ and $b=-\frac{y}{\sqrt2}.$
Thus, $x^2+y^2=2$ and we need to prove that: $$5+xy-2\sqrt2(x+y)\geq6-4\sqrt2$$ or $$2\sqrt2(2-x-y)\geq1-xy$$ or $$\frac{4\sqrt2(1-xy)}{2+x+y}\geq1-xy$$ and since $$1-xy=\frac{(x-y)^2}{2}\geq0,$$ it's enough to prove that: $$2+x+y\leq4\sqrt2,$$ which is true by C-S: $$2+x+y\leq2+\sqrt{(1+1)(x^2+y^2)}=4<4\sqrt2.$$