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

1 Answer
Sep 14, 2015

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]}