Question

If $\frac{x}{3y}=\frac{y}{2x-5y}=\frac{6x-15y}{x}$ and the expression $(-4x^2+36y-8)$ takes the maximum value for $x=m$ and $y=n$ then $m+n=$

• A. $2011$
• B. $2$
• C. $3$
• D. $17$

Answer 1

Answer: (B)

$2$

$\frac{x}{3y}=\frac{y}{2x-5y}=\frac{6x-15y}{x}$

$\Rightarrow \frac{x}{3y}=\frac{y}{2x-5y}$

$\Rightarrow 2x^2-5xy=3y^2\to \left(1\right)$

$\frac{x}{3y}=\frac{6x-15y}{x}$

$x^2=18xy-45y^2$

$x^2-18xy=-45y^2\to \left(2\right)$

$15\times \left(1\right)+\left(2\right),$

$\Rightarrow 31x^2-93xy=0$

$\Rightarrow 31x=93y\to \left(3\right)$

expression $\left(f\right)=-4x^2+36y-8$

$\Rightarrow f=-4x^2+\frac{36\times 31x}{93}-8$

$f^{\prime }=-8x+12=0$

$\therefore x=\frac{3}{2}$ & $y=\frac{1}{2}$

$\therefore x+y=2$