Minimum value too dangerous

Algebra Level 4

$433x^{2}+y^{2}+\sqrt{1729}xy+\sqrt{3}x-\sqrt{3}y+1729!$

For real values of $x$ and $y$ the above expression attains minimum value at $x=M$ and $y=N$ . And $M+N=a\sqrt{b}$ , where $a$ is positive integer and $b$ is square-free. Find $a-b$ .

The answer is 285.

1 solution

Anand Raj
Jun 5, 2019

Let the whole function be $f(x,y)$ , then the minimum/maximum occurs when $\frac{\partial f(x,y)}{\partial x}=0$ and $\frac{\partial f(x,y)}{\partial y}=0$ .

After partial differentiation, we get two linear equations in $x$ and $y$ .

$866x+\sqrt{1729}y+\sqrt{3}=0$ and $\sqrt{1729}x+2y=\sqrt{3}=0$ .

After solving these two linear equations, we get $x=\frac{-2-\sqrt{1729}}{\sqrt{3}}$ and $y=\frac{866+\sqrt{1729}}{\sqrt{3}}$ .

Thus $x+y=288\sqrt{3}$ .

Minimum value too dangerous

The answer is 285.

1 solution

0 pending reports