Answer to Question #164753 in Calculus for Balachandranaidu Babu

Question

Discuss the maxima and minima of f(x,y)=x³y²(24 –x - y).

Answer 1

$f(x,y) = x^3y^2(24-x-y) = 24x^3y^2-x^4y^2-x^3y^3$

necessary extremum condition:

Let $M_0(x_0, y_0)$ - extremum, then:

$\begin{cases} \cfrac{\partial f}{\partial x}(x_0,y_0) = 0\\\cfrac{\partial f}{\partial y}(x_0,y_0) = 0 \end{cases}$

Find first partial derivatives:

$\cfrac{\partial f}{\partial x} = 72x^2y^2 - 4x^3y^2-3x^2y^3 \\ \cfrac{\partial f}{\partial y} = \\ \begin{cases} \cfrac{\partial f}{\partial x} = 72x^2y^2 - 4x^3y^2-3x^2y^3 = 0 \\ \cfrac{\partial f}{\partial y} = 48x^3y-2x^4y-3x^3y^2=0 \\ \end{cases} \\$

Solve this system for $x,y$:

$\begin{cases} x^2y^2(72-4x-3y) = 0\\ x^2y(48x-2x^2-3y) = 0 \end{cases}\\ \begin{cases} x^2y^2 = 0 \,or\,-4x-3y+72 = 0\\ x^2y =0\, or\,-2x^2+48x-3y=0 \end{cases}$

Find the roots of first line:

$x^2y^2 = 0 \,or\,-4x-3y+72 = 0$

From second equation:

$y = 24 - \cfrac{4}{3}x$

And put it in first equation:

$24x-\cfrac{4}{3}x^2 = 0\\ -4x^2+72x=0\\ x(-4x+72) =0\\ x=0\;or \; x =18\\ y=24\; or\;y=0$

$M_1(0;24),M_2(18;0)$

Find the roots of second line:

$x^2y=0\, or\,(48x-2x^2-3y) = 0$

From second equation:

$y = -\cfrac{2}{3}x^2+16x$

And put it in first equation:

$x^2(-\cfrac{2}{3}x^2+16x)=0\\ x^3(-\cfrac{2}{3}x+16)= 0\\ x = 0\;or\; x =24\\ y=0\; or \;y=0\\ M_3(0;0), M_4(24;0)$

$\begin{cases} M_1(0;24),M_2(18;0)\\ M_3(0;0), M_4(24;0) \end{cases}$

The obtained points are the solution of two  aggregates of equations, but the necessary extremum condition says that the points must be a solution to the system. Therefore, the system has one solution - $M_3(0;0)$ . So we can check sufficient condition for extremum:

If $\cfrac{\partial^2 f}{\partial x^2}(x_0;y_0) = A, \cfrac{\partial^2 f}{\partial y^2}(x_0;y_0) = B, \cfrac{\partial^2 f}{\partial x \partial y}(x_0;y_0) = C$ ,

then :

if $AC - B^2 > 0\, and\, A>0 \,(A<0), then\, M(x_0;y_0) - minimum (maximum)$

If $AC-B^2 < 0, then\, M(x_0;y_0) -isn't\; extremum$

if $AC-B^2 = 0, \; then \, we \,can't\, say \,anything.$

$\cfrac{\partial^2 f}{\partial x^2} = 144xy^2 -12x^2y^2-6xy^3\\ \cfrac{\partial^2 f}{\partial y^2} = 48x^3-2x^4-6x^3y\\ \cfrac{\partial^2 f}{\partial x \partial y} = 144x^2y - 8x^3y-49x^2y^2$

$A(0;0) = 0\;B(0;0) = 0, \;C(0;0) = 0$

$AC-B^2 = 0$, so the point $M_3(0;0)$ is not a minimum or maximum point.