Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/271

Second Step.	${\frac {\partial ^{2}f}{\partial x^{2}}}=-6x,\qquad {\frac {\partial ^{2}f}{\partial x\partial y}}=3a,\qquad {\frac {\partial ^{2}f}{\partial y^{2}}}=-6y;$
	$\Delta ={\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial x\partial y}}\right)^{2}=36xy-9a^{2}$
Third Step.	When $x=0$ and $y=0,\quad \Delta =-9a^{2}$, and there can be neither a maximum nor a minimum at $(0,0)$.

When $x=a$ and $y=a,\quad \Delta =+27a^{2}$; and since ${\tfrac {\partial ^{2}f}{\partial x^{2}}}=-6a$, we have the conditions for a maximum value of the function fulfilled at $(a,a)$. Substituting $x=a,\quad y=a$ in the given function, we get its maximum value equal to $a^{3}$.

Illustrative Example 2. Divide $a$ into three parts such that their product shall be a maximum.

Solution	Let $x=$ first part, $y=$ second part; then $a-(x+y)=a-x-y=$ third part, and the function to be examined is
$f(x,\ y)=xy(a-x-y).$
First Step.	${\frac {\partial f}{\partial x}}=ay-2xy-y^{2}=0,\qquad {\frac {\partial f}{\partial y}}=ax-2xy-x^{2}=0.$
Solving simultaneously, we get as one pair of values $x={\tfrac {a}{3}},\quad {\tfrac {a}{3}}$. [1]
Second step.	${\frac {\partial ^{2}f}{\partial x^{2}}}=-26,\qquad {\frac {\partial ^{2}f}{\partial x\partial y}}=a-2x-2y,\qquad {\frac {\partial ^{2}f}{\partial y^{2}}}=2x;$
$\Delta =4xy-\left(a-2x-2y\right)^{2}.$
Third Step.	When $x={\tfrac {a}{3}}$ and $y={\tfrac {a}{3}},\Delta ={\tfrac {a^{2}}{3}}$; and since ${\tfrac {\partial ^{2}f}{\partial x^{2}}}=-{\tfrac {2a}{3}}$, it is seen that our product is a maximum when $x={\tfrac {a}{3}},y={\tfrac {a}{3}}$. Therefore the third part is also ${\tfrac {a}{3}}$, and the maximum value of the product is ${\tfrac {a^{3}}{27}}$.

EXAMPLES

1. Find the minimum value of $x^{2}+xy+y^{2}-ax-by$.

Ans. ${\tfrac {1}{3}}\left(ab-a^{2}-b^{2}\right)$.

2. Show that $\sin x+\sin y+\cos \left(x+y\right)$ is a minimum when $x=y={\tfrac {3\pi }{2}}$, and a maximum when $x=y={\tfrac {\pi }{6}}$.

3. Show that $xe^{y+x\sin y}$ has neither a maximum nor a minimum.

4. Show that the maximum value of ${\tfrac {\left(ax+by+c\right)^{2}}{x^{2}+y^{2}+1}}$ is $a^{2}+b^{2}+c^{2}$.

5. Find the greatest rectangular parallelepiped that can be inscribed in an ellipsoid. That is, find the maximum value of $8xyz$ (= volume) subject to the condition

${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}+{z^{2}}{c^{2}}=1.$

Ans. ${\frac {8abc}{3{\sqrt {3}}}}.$

Hint. Let $u=xyz$, and substitute the value of $z$ from the equation of the ellipsoid. This gives

$u^{2}=x^{2}y^{2}c^{2}\left(1-{\frac {x^{2}}{a^{2}}}-{\tfrac {y^{2}}{b^{2}}}\right),$

where $u$ is a function of only two variables.

1. $x=0,\quad y=0$ are not considered, since from the nature of the problem we would then have a minimum.