Question

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

Answer 1

Find `dy/dx`, find the critical numbers, test the critical numbers.

Explanation:

`4x^2 + 4xy+7y^2=15`

`8x+4y+4x dy/dx +14y dy/dx=0`

`dy/dx = (-8x-4y)/(4x+14y) = (-2(2x+y))/(2x+7y)`

`dy/dx = 0` when `y = -2x`, and is undefined when `y=-2/7x`

so we have (in the original equation):

`4x^2 + 4x(-2x)+7(-2x)^2=15`

`4x^2-8x^2+28x^2=15`

Solve for `x^2 = 15/24 = 5/8 = 10/16`

`x = +- sqrt10/4`.

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

For `x=-sqrt10/4` you should get a positive `y`
and for `x = sqrt10/4` a negative `y`.
Obviously the high point is at the positive `y`.

(Really, we also have critical points when `y=-2/7x`, so `dy/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 `+- sqrt70/4`)

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