Maximum value of $(1-a)(1-b)+(1-p)(1-q)$

Question

Given that real numbers $a,b,p,q$ satisfy $$a^2+b^2=p^2+q^2=2$$ Find the maximum value of $E=(1-a)(1-b)+(1-p)(1-q)$.

My try:

I have chosen $a=\sqrt{2}\sin x, b=\sqrt{2}\cos x, p=\sqrt{2}\sin y, q=\sqrt{2}\cos y$.

So we get $$E=2+\sin(2x)+\sin(2y)-\sqrt{2}\left(\sin x+\cos x+\sin y+\cos y\right)$$

Any help from here?

Answer 1

As stated in the comments, you could check the method used by the AOPS community, but since you need help proceeding further in the trigonometric part, I will be posting that answer :

Since x, y are independent, we can calculate max for the x containing terms and use that in the y containing terms. So the x containg terms are :

$\sin(2x) - \sqrt2(\sin x + \cos x)$

Considering $\sqrt2(\sin x + \cos x)$ Dividng by $\sqrt2$ inside bracket and multiplying outside bracet we get :

$= 2(\frac{1}{\sqrt2} \cdot \sin x + \frac{1}{\sqrt2} \cdot \cos x) = 2(\cos(\frac{\pi}{4}) \sin x + \sin(\frac{\pi}{4}) \cos x) = 2(\sin(x + \frac{\pi}{4}))$

Putting $x +\frac{\pi}{4} = t \implies x = t - \frac{\pi}{4}$ we get :

$\sin(2(t - \frac{\pi}{4})) - 2\sin t \equiv \sin(2t - \frac{\pi}{2}) - 2\sin t \equiv -\sin(\frac{\pi}{2} - 2t) - 2\sin t = -\cos(2t) - 2\sin t = - (1 - 2\sin ^2 t) - 2\sin t \equiv 2\sin ^2 t - 2\sin t - 1$

We need the max of this function, so we assume $\sin t = x \;\&\; x \in [-1,1]$

$\therefore 2x^2 - 2x - 1$ has it's minimum at $x = \frac{1}{2} (\frac{-b}{2a}$ where b is the coefficient of x and a is the coefficient of $x^2$)

Since this is an increasing quadratic equation (which is symmetric about the point where it attains it's minimum value) whichever value of x is "farthest" from when it attains minimum, we will find the function's maximum at that x for that interval (because the more you go away from the point where an increasing quadratic function attains it's minimum, the more the value of the function increases)

What I mean is $-1$ is "farther" from $\frac{1}{2}$ ("distance" $= \frac{3}{2}$) than $1$ is from $\frac{1}{2}$ ("distance" $= \frac{1}{2}$)

So putting $x = -1$ we get the maximum value of the function in the interval $[-1,1]$ which is equal to :

$2(-1)^2 - 2(-1) - 1 = 2 + 2 - 1 = 3$

Thus the maximum value of $\sin(2x) - \sqrt2(\sin x + \cos x)$ maximum value of $\sin(2y) - \sqrt2(\sin y + \cos y) = 3$

Thus maximum of the function you got is $2 + 3 + 3 = 8$

Answer 2

We can also use the method of Lagrange multipliers.

Maximize $f(x,y)=(1-x)(1-y)$ subject to $g(x,y)=x^2+y^2-2=0.$

$\frac{\partial f}{\partial x}=\lambda \frac{\partial g}{\partial x}$ and $\frac{\partial f}{\partial y}=\lambda \frac{\partial g}{\partial y}$ together gives $\frac{1-y}{1-x}=\frac xy$ and $y=x$ or $y=1-x$. So maximum of $f$ is $4$ at $x=y=-1.$

Another conclusion: $\max E=4+4=8$ where $E$ is as described in OP.

Answer 3

By AM-GM and C-S we obtain: $$(1-a)(1-b)+(1-p)(1-q)\leq\frac{(1-a)^2+(1-b)^2}{2}+\frac{(1-p)^2+(1-q)^2}{2}=$$ $$=\frac{4-2(a+b)}{2}+\frac{4-2(p+q)}{2}\leq\frac{4+2\sqrt{2(a^2+b^2)}}{2}+\frac{4+2\sqrt{2(p^2+q^2)}}{2}=8.$$ The equality occurs for $a=b=p=q=-1,$ which says that $8$ is a maximal value.

Answer 4

Consider the circle $x^2+y^2=2$. We see that $A(a,b)$ and $B(p,q)$ are two points on this circle

$(x-a)(x-p)+(y-b)(y-q)$ is the circle with $AB$ as diameter and $(1-a)(1-p)+(1-b)(1-q)$ is then the square of the length of the tangent from $(1,1)$ to this circle

Its easy to see that the maximum value is attained for the point circle at the diametrically opposite point i.e. at $(-1,-1)$ which is $8$