Minimum value of equation - Exam Edition

Given that $x+y = \sqrt{10}$ . then $(x+y)^2 = x^2 + 2xy + y^2 = 10 \implies x^2 + y^2 = 10 - 2xy$ . Now we have:

$\begin{aligned} A & = (x^4+1)(y^4+1) \\ & = x^4y^4 + x^4 + y^4 + 1 \\ & = x^4y^4 + (x^2 + y^2)^2 - 2x^2y^2 + 1 \\ & = x^4y^4 + (10-2xy)^2 - 2x^2y^2 + 1 & \small \blue{\text{Let }z =xy} \\ & = z^4 + 4z^2 - 40z + 100 - 2z^2 + 1 \\ & = z^4 + 2z^2 - 40z + 101 \end{aligned}$

To find the minimum $A(z)$ , we first find the value of $z$ such that $\dfrac {dA}{dz} =0$ and $\dfrac {d^2 A}{dz} >0$ .

$\begin{aligned} \frac {dA}{dz} & = 4 z^3 + 4z - 40 & \small \blue{\text{Putting }\frac {dA}{dz} = 0} \\ z^3 + z - 10 & = 0 \\ \implies z & = 2 & \small \blue{\text{Since }\frac {d^2A}{dz^2} > 0} \\ \implies \min (A) & = A(2) \\ & = 16+8-80+101 \\ & = \boxed{45} \end{aligned}$

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Bonus: When $z= 2$ , we have:

$\begin{aligned} xy & = 2 & \small \blue{\text{Since }x+y = \sqrt{10}} \\ x(\sqrt{10} - x) & = 2 \\ \sqrt{10} x - x^2 & = 2 \\ x^2 - \sqrt{10}x + 2 & = 0 \\ \implies x, y & = \frac {\sqrt{10} \pm \sqrt 2}2 \end{aligned}$