Find the minimum value of $7x-24y$

Question

LMNAS $25^{th}$ UGM, Indonesian

Suppose that $x,y\in\mathbb{R}$, so that :

$x^2+y^2+Ax+By+C=0$

with $A,B,C>2014$. Find the minimum value of $7x-24y$

$x^2+y^2+Ax+By+C=0$

can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$

Stuck,:>

Answer 1

Using Cauchy-Schwarz inequality, $$\left(x+\frac{A}{2}\right)^2+\left(y+\frac{B}{2}\right)^2+C-\left(\frac{A}{2}\right)^2-\left(\frac{B}{2}\right)^2=0\\ \implies 7x-24y=7\left(x+\frac A2\right)-24\left(y+\frac B2\right)+\frac{-7A+24B}{2}\\ \ge -\sqrt{7^2+(-24)^2} \cdot \sqrt{\left(x+\frac A2\right)^2+\left(y+\frac B2\right)^2}+ \frac{-7A+24B}{2}\\ =-25 \frac{1}{2} \sqrt{A^2+B^2-4C} + \frac{-7A+24B}{2}\\ =\frac 12 \left(-25\sqrt{A^2+B^2-4C}-7A+24B\right).\blacksquare$$

Answer 2

Since $$ x^2+Ax+\left( \frac{A}{2} \right) ^2+y^2+By+\left( \frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C\\ \left( x+\frac{A}{2} \right) ^2+\left( y+\frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C $$ Let $\displaystyle r=\sqrt{\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C}$, we can set $x=r\cos \theta -\dfrac{A}{2},y=r\sin \theta -\dfrac{B}{2}$, then $$ \begin{aligned} 7x-24y&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} (7 \cos \theta-24 \sin \theta)-7 A+24 B\right)\\ &=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} \sqrt{7^2+24^2}\sin(\theta+\phi)-7 A+24 B\right)\\ &\geq \frac{1}{2} \left(25\sqrt{A^2+B^2-4 C} -7 A+24 B\right) \end{aligned} $$ The second step uses The auxiliary Angle formula.

Answer 3

Hint:

Let $7x-24y=z\iff x=?$

Replace the value of $x$ in $$x^2+y^2+Ax+By+C=0$$ to form a Quadratic Equation in $y$

As $y$ is real, the discriminant must be $\ge0$