# How to find critical points and the highest point on the graph of 4x^2 + 4xy+7y^2=15?

Sep 14, 2015

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, find the critical numbers, test the critical numbers.

#### Explanation:

$4 {x}^{2} + 4 x y + 7 {y}^{2} = 15$

$8 x + 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 14 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 8 x - 4 y}{4 x + 14 y} = \frac{- 2 \left(2 x + y\right)}{2 x + 7 y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ when $y = - 2 x$, and is undefined when $y = - \frac{2}{7} x$

so we have (in the original equation):

$4 {x}^{2} + 4 x \left(- 2 x\right) + 7 {\left(- 2 x\right)}^{2} = 15$

$4 {x}^{2} - 8 {x}^{2} + 28 {x}^{2} = 15$

Solve for ${x}^{2} = \frac{15}{24} = \frac{5}{8} = \frac{10}{16}$

$x = \pm \frac{\sqrt{10}}{4}$.

Now substitute in the original equation and solve for $y$.

For $x = - \frac{\sqrt{10}}{4}$ you should get a positive $y$
and for $x = \frac{\sqrt{10}}{4}$ a negative $y$.
Obviously the high point is at the positive $y$.

(Really, we also have critical points when $y = - \frac{2}{7} x$, so $\frac{\mathrm{dy}}{\mathrm{dx}}$ does not exist. But the tangent lines at those points are vertical, so they are not high or low points on the ellipse. The $x$ values are $\pm \frac{\sqrt{70}}{4}$)

Here is the graph:
graph{4x^2 + 4xy+7y^2=15 [-4.385, 4.384, -2.19, 2.195]}