Question

maximum or minimum of f(x, y) = x³+ y³– 63(x + y) + 12xy

Answer 1

The two variables function is given by

$F(x,y) = x^3 + y^3 - 63(x+y) + 12xy$

The first order partial derivatives of this function are

$F_x = 3x^2 - 63 + 12y \, \, and \,\, F_y = 3y^2 - 63 + 12x$

The second order partial derivatives of this function are

$F_{xx} = 6x \,\,\, , \,\,\, F_{yy} = 6y \,\,\, and \,\,\, F_{xy} = 12$

To find critical points of this function, we have to set $F_x = 0 \,\,\, and \,\,\, F_y = 0$

Hence:

$F_x = 0 \\ \Rightarrow 3x^2 - 63 + 12y = 0 \\ \Rightarrow 12y = 63 - 3x^2$

$\Rightarrow 4y = 21 - x^2 \\ \Rightarrow y = \frac{21 - x^2}{4}$ (Equation 1)

And:

$F_y = 0 \\ \Rightarrow 3y^2 - 63 + 12x = 0 \\ \Rightarrow y^2 + 4x - 21 = 0$

$\Rightarrow \frac{(21 - x^2)^2}{16} + 4x - 21 = 0 \hspace{1 cm} \left[ \because y = \frac{21 - x^2}{4} \right]$

$\Rightarrow 441 - 42x^2 + x^4 + 64x - 336 = 0 \\ \Rightarrow x^4 - 42x^2 + 64x + 105 = 0 \\$

Using a graphing calculator, we get the solutions of this polynomial equation and which are:

$x = -7 \,\,\, , \,\,\, x = -1 \,\,\, , \,\,\, x = 3 \,\,\, and \,\,\, x = 5$

Inserting $x = -7$ into the (Equation 1), we have:

$y = \frac{21 - (-7)^2}{4} \\ \Rightarrow y = -7$

Inserting $x = -1$ into the (Equation 1), we have:

$y = \frac{21 - (-1)^2}{4} \\ \Rightarrow y = 5$

Inserting $x = 3$ into the (Equation 1), we have:

$y = \frac{21 - 3^2}{4} \\ \Rightarrow y = 3$

Inserting $x = 5$ into the (Equation 1), we have:

$y = \frac{21 - 5^2}{4} \\ \Rightarrow y = -1$

Hence the critical points of this function are

$(-7, -7) \,\,\, , \,\,\, (-1,5) \,\,\, , \,\,\, (3,3) \,\,\, and \,\,\ (5, -1)$

Now,

to classify the critical points and obtain the relative extrema values of this function, we will use the following value of  $D$

$D = F_{xx} F_{yy} - (F_{xy})^2 \\ \,\,\,\,\,\, = (6x)(6y) -(12)^2 \\ \,\,\,\,\,\, = 36xy - 144$

At $(-7,-7)$:

$D = 36(-7)(-7) - 144 \\ \,\,\,\,\,\,\, = 1620 > 0$

and:

$F_{xx} (-7,-7) \,= 6(-7) \\ \hspace{1 cm} \hspace{1 cm} = -42 < 0$

At $(-1,5)$:

$D = 36(-1)(5) - 144 \\ \,\,\,\,\,\,\, = -324 < 0$

At $(3,3)$:

$D = 36(3)(3) - 144 \\ \,\,\,\,\,\,\, = 180> 0$

$F_{xx} (3,3) \,= 6(x) \\ \hspace{1 cm} \hspace{1 cm} = 18 > 0$

At $(5,-1)$:

$D = 36(5)(-1) - 144 \\ \,\,\,\,\,\,\, = -324 < 0$

So we can say that:

The saddle points of this function are

$(5,-1) \,\,\,\, and \,\,\,\, (-1,5)$

The relative maximum of this function exists at $(-7,-7)$

and

the relative minimum of this function exists at $(3,3)$

The relative maximum value of the function is

$F(-7,-7) = (-7)^3 + (-7)^3 - 63(-7-7)$ $+ 12(-7)(-7)$

$\hspace{1 cm} \,\,\,\,\,\,\,\,\,\,\,\,\, = 784$

The relative minimum value of the function is

$F(3,3) = 3^3 + 3^3 - 63(3+3) + 12(3)(3)$

$\hspace{1cm} \,\,= -216$