Question

2) The point (1,2) is: a. a local maximum for f b. a local minimum for f c. a saddle point for f a Ο Ο Ο b Exercise 6. Double integral in rectangular coordinates (10 pts+10 pts) Let I = secx dydx.

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Abdul-Rahim Taysir

Answer 1

Solution:-

Given that

2)

Since given

$f(x,y)=2x^2-4x+y^2-4y+1$ .........(1)

so,

$\frac{\partial f}{\partial x}=4x-4$

and

Әf ду 2y – 4

for critical point

0 д
and $\frac{\partial f}{\partial y}=0$

$4x-4=0$,   $2y-4=0$

$x=1$, $y=2$

So, critical point of function (1) is (1, 2)

So, the number of critical points of f is 1.

Since a point $(x_0,y_0)$ is

(i) maximum point if

(ii) minimum point if

(iii) and neither maximum nor minimum if $AC-B^2<0$

where

$A=\frac{\partial^2 f}{\partial x^2}$,  $B=\frac{\partial^2 f}{\partial x \partial y}$ and $B=\frac{\partial^2 f}{\partial y^2}$

Now

$A=\frac{\partial^2 f}{\partial x^2}=4,$ B = 0,  $C=\frac{\partial^2 f}{\partial y^2}=2$

So,

$AC-B^2=B-0=B>0$

where $A=4>0$

So, critical point (1, 2) is a local minimum for f.

Option b is correct

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