Question

At what point the function f(x, y) = x2 + 2xy + 3y2 + 4x + 6y
has minimum value?​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given function is

$\rm :\longmapsto\:f = {x}^{2} + 2xy + {3y}^{2} + 4x + 6y$

On differentiating partially w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{\partial f}{\partial x} = 2x + 2y + 4$

On differentiating partially w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{\partial f}{\partial y} = 2x + 6y + 6$

Now, for critical points,

$\rm :\longmapsto\:\dfrac{\partial f}{\partial x} = 0 \: \: and \: \: \dfrac{\partial f}{\partial y} = 0$

$\rm :\longmapsto\:2x + 2y + 4 = 0$

$\rm :\longmapsto\:x + y + 2 = 0 - - - (2)$

and

$\rm :\longmapsto\:2x + 6y + 6 = 0$

$\rm :\longmapsto\:x + 3y + 3 = 0 - - - (3)$

On Subtracting equation (2) from (3), we get

$\rm :\longmapsto\:2y + 1 = 0$

$\rm :\implies\:y = - \dfrac{1}{2}$

On substituting the value of y in equation (2), we get

$\rm :\longmapsto\:x - \dfrac{1}{2} + 2 = 0$

$\rm :\longmapsto\:x + \dfrac{3}{2} = 0$

$\rm :\implies\:x = - \dfrac{3}{2}$

Hence,

$\rm :\longmapsto\:Critical \: point \: is \: \bigg( - \dfrac{3}{2} , \: - \dfrac{1}{2} \bigg)$

Now to check whether this point is of maxima or minima or saddle point or point of inflection.

$\rm :\longmapsto\:a = \dfrac{ {\partial }^{2} f}{ {\partial x}^{2} } = 2$

$\rm :\longmapsto\:c = \dfrac{ {\partial }^{2} f}{ {\partial y}^{2} } = 6$

$\rm :\longmapsto\:b = \dfrac{ {\partial }^{2} f}{ {\partial x \: \partial y} } = 2$

So,

$\rm :\longmapsto\:a = 2$

$\rm :\longmapsto\:b = 2$

$\rm :\longmapsto\:c = 6$

Now, Consider,

$\rm :\longmapsto\:ac - {b}^{2}$

$\rm \: = \: \: 2 \times 6 - {2}^{2}$

$\rm \: = \: \: 12 - 4$

$\rm \: = \: \: 8$

Thus, we have now

$\bf :\longmapsto\:ac - {b}^{2} > 0 \: \: \: and \: \: \: a > 0$

Hence,

$\bf :\longmapsto\:f \: is \: minimum \: at \: \bigg( - \dfrac{3}{2} , \: - \dfrac{1}{2} \bigg)$

Additional Information :-

Now, Maxima and minima for 2 variables.

$\rm :\longmapsto\:ac - {b}^{2} > 0, \: a > 0, \: f \: is \: minimum.$

$\rm :\longmapsto\:ac - {b}^{2} > 0, \: a < 0, \: f \: is \: maximum.$

$\rm :\longmapsto\:ac - {b}^{2} = 0, \: f \: has \:a \: saddle \: point.$