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

Question

Evelyn Agard · Accepted Answer

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

#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]}

Christopher Allers · Answer

undefined To find critical points on the graph of $4x^2 + 4xy + 7y^2 = 15$, we first need to take partial derivatives with respect to $x$ and $y$ and then solve the resulting system of equations.

Taking the partial derivative with respect to $x$ gives us: $8x + 4y = 0$, and with respect to $y$ gives us: $4x + 14y = 0$.

Solving the system of equations simultaneously, we find that $x = -\frac{7}{3}$ and $y = \frac{7}{6}$.

Substituting these values back into the original equation, we find that the critical point is $\left(-\frac{7}{3}, \frac{7}{6}\right)$.

To find whether this critical point corresponds to a maximum or minimum, we can use the second derivative test.

The second partial derivatives with respect to $x$ and $y$ are $8$ and $14$ respectively.

The determinant of the Hessian matrix ($H$) is $D = (8)(14) - (4)^2 = 112 - 16 = 96$.

Since $D > 0$ and the second partial derivative with respect to $x$ is positive, the critical point corresponds to a minimum.

Therefore, the highest point on the graph corresponds to the critical point $\left(-\frac{7}{3}, \frac{7}{6}\right)$.