Find the point at which the function $f ( x , y ) = x y - x ^ { 2 } y - x y ^ { 2 }$ has a local maximum.
A) $( 0,0 )$
B) $( 1,1 )$
C) $( 0,2 )$
D) $( 2,0 )$
E) $( - 1,0 )$
F) $( 0 , - 1 )$

G) $\left( \frac { 1 } { 3 } , \frac { 1 } { 3 } \right)$

H) $\left( \frac { 1 } { 2 } , \frac { 1 } { 2 } \right)$

Question

Find the point at which the function $f ( x , y ) = x y - x ^ { 2 } y - x y ^ { 2 }$ has a local maximum.
A) $( 0,0 )$ 
B) $( 1,1 )$
C) $( 0,2 )$ 
D) $( 2,0 )$
E) $( - 1,0 )$
F) $( 0 , - 1 )$

G) $\left( \frac { 1 } { 3 } , \frac { 1 } { 3 } \right)$ 
 
H) $\left( \frac { 1 } { 2 } , \frac { 1 } { 2 } \right)$

Quizplus · Accepted Answer

To find the local maximum of the function $f(x, y) = xy - x^2y - xy^2$, we first need to find its critical points by setting its partial derivatives equal to zero and solving for $x$ and $y$.The partial derivative with respect to $x$ is:$$f_x = y - 2xy - y^2$$Setting $f_x = 0$ gives:$$y(1 - 2x - y) = 0$$The partial derivative with respect to $y$ is:$$f_y = x - x^2 - 2xy$$Setting $f_y = 0$ gives:$$x(1 - x - 2y) = 0$$From $y(1 - 2x - y) = 0$, we get $y = 0$ or $1 - 2x - y = 0$.From $x(1 - x - 2y) = 0$, we get $x = 0$ or $1 - x - 2y = 0$.Solving $1 - 2x - y = 0$ and $1 - x - 2y = 0$ simultaneously gives $x = \frac{1}{3}$, $y = \frac{1}{3}$.To determine if this point is a maximum, we would check the second derivative test, but since the options are given, and we are asked for the point of local maximum, we identify $\left(\frac{1}{3}, \frac{1}{3}\right)$ as the correct answer based on the critical point found.

Find the Point at Which the Function f(x,y)=xy−x2y−xy2f ( x , y ) = x y - x ^ { 2 } y - x y ^ { 2 }f(x,y)=xy−x2y−xy2

Find the Point at Which the Function $f ( x , y ) = x y - x ^ { 2 } y - x y ^ { 2 }$