Find the minimum value of $f(a,b) = 5 - 2ab -4(a - b)$ subject to $a^2+b^2 = 1$ without calculus

Question

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 $f(a,b) = 5 - 2ab - 4(a-b)$.

Answer 1

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: $$2+x+y\leq2+\sqrt{(1+1)(x^2+y^2)}=4<4\sqrt2.$$

Answer 2

I would recommend you to notice that:

\begin{align*} 5 - 2ab - 4(a - b) & = 4 + (a^{2} + b^{2}) - 2ab - 4(a - b)\\\\ & = (a^{2} - 2ab + b^{2}) - 4(a - b) + 4\\\\ & = (a - b)^{2} - 4(a - b) + 4\\\\ & = (a - b - 2)^{2}\\\\ & = (\cos(\theta) - \sin(\theta) - 2)^{2}\\\\ & = \left(\sqrt{2}\cos\left(\theta + \pi/4\right) - 2\right)^{2}\\\\ & = |\sqrt{2}\cos(\theta + \pi/4) - 2|^{2} \end{align*}

where $a = \cos(\theta)$ and $b = \sin(\theta)$. Moreover, we do also have that: \begin{align*} |\cos(\theta + \pi/4)| \leq 1 & \Longleftrightarrow |\sqrt{2}\cos(\theta + \pi/4)| \leq \sqrt{2}\\\\ & \Longleftrightarrow -\sqrt{2} \leq \sqrt{2}\cos(\theta) + \pi/4) \leq \sqrt{2}\\\\ & \Longleftrightarrow -2 - \sqrt{2} \leq \sqrt{2}\cos(\theta + \pi/4) - 2\leq \sqrt{2} - 2\\\\ & \Longleftrightarrow 2 - \sqrt{2} \leq |\sqrt{2}\cos(\theta + \pi/4) - 2| \leq 2 + \sqrt{2} \end{align*}

Consequently, the minimum value corresponds to $(2 - \sqrt{2})^{2} = 6 - 4\sqrt{2}$.

Hopefully this helps!

Answer 3

Geometrically the problem is equivalent to minimizing the distance from the circle to the line plus the circle radius since it's a constant. Thus, it's sufficient to find the point of intersection of line perpendicular to $y=x-2$ and going thru the center of the unit circle with the circle $x^2+y^2=1$. Such line is $y=-x$ and the result follows.