Answer
(a) We solve the equations ${f_x} = 0$ and ${f_y} = 0$ and obtain the set of critical points of $f$, which is the parabola $y = {x^2}$.
But the discriminant $D = 0$ at the critical points. Hence, the Second Derivative Test fails for these points.
(b) We show that for fixed $b$, the function $g\left( x \right) = f\left( {x,b} \right)$ is concave up for $x>0$ with a critical point at $x = {b^{1/2}}$.
(c) We show that the critical point $\left( {x,y} \right) = \left( {{b^{1/2}},b} \right)$ is the local minimum. So, we conclude that $f\left( {a,b} \right) \ge f\left( {{b^{1/2}},b} \right) = 0$ for all $a,b \ge 0$.
Work Step by Step
(a) We have $f\left( {x,y} \right) = \frac{1}{3}{x^3} + \frac{2}{3}{y^{3/2}} - xy$.
The partial derivatives are
${f_x} = {x^2} - y$, ${\ \ \ }$ ${f_y} = {y^{1/2}} - x$
${f_{xx}} = 2x$, ${\ \ }$ ${f_{yy}} = \frac{1}{2}{y^{ - 1/2}}$, ${f_{xy}} = - 1$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = {x^2} - y = 0$, ${\ \ \ }$ ${f_y} = {y^{1/2}} - x = 0$
Both equations lead to the solution $y = {x^2}$. Hence, the set of critical points of $f$ is the parabola $y = {x^2}$.
The discriminant $D = {f_{xx}}{f_{yy}} - {f_{xy}}^2$ at the critical points becomes
$D = 2x\cdot\frac{1}{2}{\left( {{x^2}} \right)^{ - 1/2}} - {\left( { - 1} \right)^2} = 0$
Hence, the Second Derivative Test fails for these points.
(b) For fixed $b$, we get a single variable function
$g\left( x \right) = f\left( {x,b} \right) = \frac{1}{3}{x^3} + \frac{2}{3}{b^{3/2}} - bx$.
The derivatives are
$g'\left( x \right) = {x^2} - b$, ${\ \ \ }$ $g{\rm{''}}\left( x \right) = 2x$
The critical points of $g$ is obtained by solving $g'\left( x \right) = {x^2} - b = 0$. Thus, the critical point is at $x = \pm {b^{1/2}}$.
Recall from Section 4.5, $g$ is concave up if $g' > 0$ and $g{\rm{''}} > 0$. Hence, the function $g\left( x \right) = f\left( {x,b} \right)$ is concave up for $x > 0$ with a critical point at $x = {b^{1/2}}$.
(c) Recall from part (a): the set of critical points of $f$ is the parabola $y = {x^2}$.
Since $g\left( x \right) = f\left( {x,b} \right)$ is concave up for $x>0$ with a critical point at $x = {b^{1/2}}$. This implies that the critical point $\left( {x,y} \right) = \left( {{b^{1/2}},b} \right)$ is the local minimum. The minimum value of $f$ is $f\left( {{b^{1/2}},b} \right)$:
$f\left( {{b^{1/2}},b} \right) = \frac{1}{3}{b^{3/2}} + \frac{2}{3}{b^{3/2}} - b\left( {{b^{1/2}}} \right) = 0$
Thus, we conclude that $f\left( {a,b} \right) \ge f\left( {{b^{1/2}},b} \right) = 0$ for all $a,b \ge 0$.
