Calculus problem #1769

We have $\begin{aligned} \frac{d}{dx}(0) &= \frac{d}{dx}(-3x^2) + \frac{d}{dx}(2y^2) + \frac{d}{dx}(-5xy) + \frac{d}{dx}(300) \\ 0 &= -6x + 2\frac{d}{dy}(y^2)\frac{dy}{dx} - 5y - 5x\frac{dy}{dx} + 0 \\ &= -6x + 4y\frac{dy}{dx} - 5y - 5x\frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{6x + 5y}{4y - 5x} \\ \end{aligned}$

Thus $\left[\frac{dy}{dx}\right]_{(7,9)} = \frac{6(7) + 5(9)}{4(9) - 5(7)} = 87$ .

$\begin{array} {rll} - 3x^2 + 2y^2 - 5xy + 300 & = 0 & \small \color{#3D99F6} {\text{Differentiate both sides.}} \\ \Rightarrow - 6x + 4y\dfrac{dy}{dx} - 5y - 5x\dfrac{dy}{dx} + 0 & = 0 & \small \color{#3D99F6} {\text{Substitute in } x = 7 \text{ and } y = 9} \\ \Rightarrow -42 + 36\dfrac{dy}{dx} - 45 - 35 \dfrac{dy}{dx} & = 0 & \small \color{#3D99F6} {\text{Rearranging, we have...}} \\ \left[ \dfrac{dy}{dx} \right]_{(7,9)} & = \boxed{87} \end{array}$