Question

$x^2+y^2+xy=1$ for all $x,y\in R,$ the minimum value of $x^{3}y+x{y}^3+4$ is
(correct answer + 1, wrong answer - 0.25)

• A. $1$
• B. $3$
• C. $4$
• D. $2$

Answer 1

The correct option is D $2$

Let $xy=t$
$x^2+y^2+xy=1\Rightarrow x^2+y^2=1-t\ge 0$
$\Rightarrow t\le 1\forall t\in R$

$x^2+y^2+xy=1\Rightarrow (x+y{)}^2=1+t\ge 0$
$\Rightarrow t\ge -1\forall t\in R$
$\Rightarrow -1\le t\le 1$

$f\left(t\right)=t(x^2+y^2)+4=t(1-t)+4=t-t^2+4=-[t^2-t-4]$
$f\left(t\right)=-[{(t-\frac{1}{2})}^2-\frac{17}{4}]=\frac{17}{4}-{(t-\frac{1}{2})}^2$
$\therefore f(t{)}_{min}=\frac{17}{4}-{(-1-\frac{1}{2})}^2=2$
which is occur when $t=xy=-1$