Question

Find the critical point of the function f(x,y)=x^{2}+y^{2}+4xy+48x

c=

Use the Second Derivative Test to determine whether the point is
A. a saddle point
B. a local maximum
C. test fails
D. a local minimum

Answer 1

anyone? i need help

Answer 2

Critical points are found by finding when \(f_x(x,y)= 0\) and \(f_y(x,y)=0\). In this instance, \(f_x(x,y) = 2x+4y+48\) and \(f_y(x,y)=2y+4x\).

Hence, we have to solve the system
\[\left\{\begin{aligned} 2x+4y &= -48 \\ 2y+4x &= 0\end{aligned}\right.\] The second equation implies that \(y=-2x\). Substituting this into the first equation gives us \(2x-8x=-48 \implies -6x = -48 \implies x=8\). Thus, \(y=-16\) and therefore the critical point is \((8,-16)\).

Do you know how to apply the second partials test and take things from here?