Question

u=2a²xy - 3ax²y-9y³+ x³y + xy³ find maxima and minima point by partial differentiation​

Answer 1

Hope it helps.!!!!!!!!

Answer 2

$\huge\mathcal\colorbox{black}{{\color{red}{❥Solution}}}$

$3x + 2y = - 4 - - - - - (1) \\ 2x + 5y = 1 - - - - - (2)$

⌧ Find Point of Intersection of these 2 lines

Do 2×(1)-3×(2)

$6x + 4y = - 8 - - - - - (3) \\ 6x + 5y = 3 - - - - - (4) \\ - - - - - - - - - - - - \\ - 11y = - 11 \\ y = 1$

$\bf\red {Substitute in (1) :-}$

$3x + 2(1) = - 4 \\ = > 3x = - 4 - 2 \\ = > 3x = - 6 \\ = > x = - 2$

.

. . Point of Intersection of the line

$3x + 2y + 4 = 0 \: and \: 2x + 5y - 1 = 0 \: is( - 2 \: and \: 1)$

$y = mx + c$

Substitute

$x = - 2 \: amd \: y \: = 1 \\ \\ = > 1 = - 2m + c \\ = > c = 1 + 2m$

.

. .The required equation of straight line is form :-

$y = mx + 1 + 2m - - - - - (5)$

⌧ The perpendicular distance (or simply distance ) 'd' of a point P (x1, y1) from Ax+By+C = 0 is given by.

$\bf\red {Given :-}$

$(x1, y1) = ( - 2 , 1) \: andd = 2 \\ \\ = > (4m + {2})^{2} = 2( {m}^{2} + 1) \\ = > 16 {m}^{2} + 16m + 4 = 2 {m}^{2} + 2 \\ = > 14 {m}^{2} + 16m + 2 = 0 \\ = > 7 {m}^{2} + 8m + 1 = 0$

Factories the equation

$7 {m}^{2} + 7m + m + 1 = 0 \\ = > 7m(m + 1) + 1(m + 1) = 0 \\ = > m = - 1 \: and \: m = \frac{ - 1}{7}$

Now substitute m = -1 in (5)

$= > y = ( - 1)x + 1 + 2( - 1) \\ = > y = ( - 1)x + 1 + 2( - 1) \\ = > y = - x + 1 - 2 \\ y = - 1 - x$