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



When $$x =a$$ and $$y = a, \quad \Delta = + 27 a^2$$; 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 $$a^3$$. Divide $$a$$ into three parts such that their product shall be a maximum.
 * Second Step.
 * align="center" | $$\frac{\partial^2 f}{\partial x^2} = -6x, \qquad \frac{\partial^2 f}{\partial x \partial y} = 3a, \qquad \frac{\partial^2 f}{\partial y^2} = -6y;$$
 * align="center" | $$\Delta = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 = 36xy - 9a^2$$
 * Third Step.
 * align="left"| When $$x = 0$$ and $$ y = 0, \quad \Delta = -9 a^2$$, and there can be neither a maximum nor a minimum at $$(0, 0)$$.
 * }
 * Third Step.
 * align="left"| When $$x = 0$$ and $$ y = 0, \quad \Delta = -9 a^2$$, and there can be neither a maximum nor a minimum at $$(0, 0)$$.
 * }


 * {| style="width:100%"


 * align="left"| Solution
 * align="left"| Let $$x = $$ first part, $$y = $$ second part; then $$a - (x+y) = a - x- y = $$ third part, and the function to be examined is
 * colspan="2" align="center"|$$f(x,\ y) = xy(a-x-y).$$
 * valign="top" | First Step.
 * align="center"| $$ \frac{\partial f}{\partial x} = ay - 2xy - y^2 =0, \qquad \frac{\partial f}{\partial y} = ax - 2xy -x^2 =0.$$
 * colspan="2" align="left" |Solving simultaneously, we get as one pair of values $$x = \tfrac{a}{3}, \quad \tfrac{a}{3}$$.
 * valign="top" |Second step.
 * align="center"|$$\frac{\partial^2 f}{\partial x^2} = -26, \qquad \frac{\partial^2 f}{\partial x \partial y} = a - 2x - 2y, \qquad \frac{\partial^2 f}{\partial y^2} = 2x;$$
 * colspan="2" align="center"| $$\Delta = 4xy - \left( a -2x - 2y \right)^2.$$
 * valign="top" | Third Step.
 * align="left"| 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 $$\tfrac{a^3}{27}$$.
 * }
 * colspan="2" align="center"| $$\Delta = 4xy - \left( a -2x - 2y \right)^2.$$
 * valign="top" | Third Step.
 * align="left"| 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 $$\tfrac{a^3}{27}$$.
 * }
 * align="left"| 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 $$\tfrac{a^3}{27}$$.
 * }

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 $$8 xyz$$ (= volume) subject to the condition

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

$$u^2 = x^2y^2c^2\left( 1 - \frac{x^2}{a^2} - \tfrac{y^2}{b^2} \right),$$ where $$u$$ is a function of only two variables.