## Thomas' Calculus 13th Edition

Local maximum of $f(\dfrac{2}{3},\dfrac{4}{3})=0$
Given: $f_x(x,y)=2y-10x+4=0, f_y(x,y)=2x-4y+4=0$ Simplify the given two equations. This implies that $x=\dfrac{2}{3},y=\dfrac{4}{3}$ Thus, the critical point: $(\dfrac{2}{3},\dfrac{4}{3})$ In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. 1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum. 2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum. 3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum and local maximum or, a saddle point. $D=f_{xx}f_{yy}-f^2_{xy}=36$ and $D=36 \gt 0$ and $f_{xx}(\dfrac{2}{3},\dfrac{4}{3})=-10 \lt 0$ Hence, Local maximum of $f(\dfrac{2}{3},\dfrac{4}{3})=0$