Answer
(a) There is only one critical point at $\left( {\frac{1}{3},\frac{1}{3}} \right)$.
The extreme value is $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$.
(b) The extreme values of $f$ on the bottom edge is $g\left( {\frac{1}{2}} \right) = f\left( {\frac{1}{2},0} \right) = \frac{1}{4}$.
(c)
1. On the top edge of the square, $y=2$.
The extreme value of $f$ along $y=2$ is $f\left( { - \frac{1}{2},2} \right) = - \frac{7}{4}$.
2. On the left edge of the square, $x=0$.
The extreme value of $f$ along $x=0$ is $f\left( {0,\frac{1}{2}} \right) = \frac{1}{4}$.
3. On the right edge of the square, $x=2$.
The extreme value of $f$ along $x=2$ is $f\left( {2, - \frac{1}{2}} \right) = - \frac{7}{4}$.
(d) The largest value of $f$ among the values computed in (a), (b), and (c) is $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$.
The maximum of $f$ is $\frac{1}{3}$ which occurs at the critical point $\left( {\frac{1}{3},\frac{1}{3}} \right)$.
Work Step by Step
(a) We have $f\left( {x,y} \right) = x + y - {x^2} - {y^2} - xy$.
The partial derivatives are
${f_x} = 1 - 2x - y$, ${\ \ }$ ${f_y} = 1 - 2y - x$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 1 - 2x - y = 0$, ${\ \ }$ ${f_y} = 1 - 2y - x = 0$
From the first equation we get $y=1-2x$.
Substituting it in the second equation gives
$1 - 2 + 4x - x = 0$
So, the solutions are $x = \frac{1}{3}$ and $y = \frac{1}{3}$.
There is only one critical point at $\left( {\frac{1}{3},\frac{1}{3}} \right)$. So, the extreme value $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$.
(b) On the bottom edge of the square, $y=0$.
Write $g\left( x \right) = f\left( {x,0} \right) = x - {x^2}$.
Find the extreme values of $f$ on the bottom edge.
The critical points of $g$ can be found by solving the equation $g'\left( x \right) = 0$.
$g'\left( x \right) = 1 - 2x = 0$
$x = \frac{1}{2}$
So, the extreme values of $f$ on the bottom edge is $g\left( {\frac{1}{2}} \right) = f\left( {\frac{1}{2},0} \right) = \frac{1}{4}$.
(c)
1. On the top edge of the square, $y=2$.
Write $h\left( x \right) = f\left( {x,2} \right) = - 2 - {x^2} - x$.
Find the extreme values of $f$ on the top edge.
The critical points of $h$ can be found by solving the equation $h'\left( x \right) = 0$.
$h'\left( x \right) = - 2x - 1 = 0$
$x = - \frac{1}{2}$
So, the extreme value of $f$ along $y=2$ is $h\left( { - \frac{1}{2}} \right) = f\left( { - \frac{1}{2},2} \right) = - \frac{7}{4}$.
2. On the left edge of the square, $x=0$.
Write $m\left( y \right) = f\left( {0,y} \right) = y - {y^2}$.
Find the extreme values of $f$ on the left edge.
The critical points of $m$ can be found by solving the equation $m'\left( y \right) = 0$.
$m'\left( y \right) = 1 - 2y = 0$
$y = \frac{1}{2}$
So, the extreme value of $f$ along $x=0$ is $m\left( {\frac{1}{2}} \right) = f\left( {0,\frac{1}{2}} \right) = \frac{1}{4}$.
3. On the right edge of the square, $x=2$.
Write $n\left( y \right) = f\left( {2,y} \right) = - 2 - y - {y^2}$.
Find the extreme values of $f$ on the right edge.
The critical points of $n$ can be found by solving the equation $n'\left( y \right) = 0$.
$n'\left( y \right) = - 1 - 2y = 0$
$y = - \frac{1}{2}$
So, the extreme value of $f$ along $x=2$ is $n\left( { - \frac{1}{2}} \right) = f\left( {2, - \frac{1}{2}} \right) = - \frac{7}{4}$.
(d) The largest value of $f$ among the values computed in (a), (b), and (c) is $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$.
Hence, the maximum of $f$ is $\frac{1}{3}$ which occurs at the critical point $\left( {\frac{1}{3},\frac{1}{3}} \right)$.
