Question

F
For the function f(x,y) = 2xy - 5x2 - 2y2 + 4x + 4y - 4 the critical point is
is a
point of
a) Maxima
b) Minima
c) Saddle point
d) None of these​

Answer 1

Answer:

saddle point

Step-by-step explanation:

good morning

Answer 2

Answer:

The critical point is a point of maxima.

Step-by-step explanation:

Given a function, $f(x,y) = 2xy - 5x^{2} - 2y^{2} + 4x + 4y - 4$

Taking the partial derivative with respect to x and y,

$f_x(x,y) = 2y - 10x + 4$  

and  $f_y(x,y) = 2x - 4y + 4$

To find the critical point, solve both the equations for x and y.

Critical point is the interior point of $f(x,y)$ where $f_x = f_y = 0$ or $f_x$ or $f_y$ fails to exist.

$2y-10x+4=0$                            ...(1)

$2x - 4y + 4=0$                             ...(2)

Multiplying equation (1) with 2 and adding to equation (2),

$2(2y-10x+4)+(2x - 4y + 4)=0$

$\implies 4y-20x+8+2x - 4y + 4=0$

$\implies -18x=-12\implies x=\frac{12}{18} =\frac{2}{3}$

Substituting the value of x in equation (1).

$2y-10\times \frac{2}{3} +4=0$

$\implies 2y- \frac{8}{3} =0\implies 6y=8\implies y=\frac{8}{6}= \frac{4}{3}$

Therefore, $\big(\frac{2}{3}, \frac{4}{3} \big)$ is the critical point.

To find the nature of the critical point, apply the second derivative test.

Second derivative test: The nature of $f (a, b)$ can be tested, if the first-and second-order partial derivatives of $f$ are continuous at a critical point (a, b) and $f_x (a, b) = f_y (a, b) = 0$.

1. Local maximum  $f_{xx} < 0$ and $f_{xx} f_{yy} -f_{xy}^2 > 0$ at (a, b)
2. Local minimum when $f_{xx} > 0$ and $f_{xx} f_{yy} -f_{xy}^2 > 0$ at (a, b)
3. At saddle point when $f_{xx} f_{yy} -f_{xy}^2 < 0$ at (a, b)
4. Test is inconclusive if $f_{xx} f_{yy} -f_{xy}^2 = 0$ at (a, b)

Taking the second order partial derivative with respect to x and y,

$f_{xx}(x,y) = - 10$  

$f_{yy}(x,y) = - 4$

$f_{xy}(x,y) = 2$  

Thus at the critical point $\big(\frac{2}{3}, \frac{4}{3} \big)$ ,

$f_{xx}(x,y) = - 10 < 0$ and

$f_{xx} f_{yy} -f_{xy}^2 =(-10) (-4) -2^2=36 > 0$

Therefore, critical point is local maximum.

So, the correct answer is option a.