Find all local extreme values of the given function and identify each as a local

Question

Find all local extreme values of the given function and identify each as a local maximum, local minimum, or saddle point. 9) fox, y)-36x2+18xy+9y2 9) A) f(18, 18)-9720, local maximum; f(0, 0)-0, local minimum B) f(18, 18)-9720, local maximum C) f(o, 0)-0, local minimum D) f(6,3)-1701, saddle point, f(3, 6)-972, saddle point Find the extreme values of the function subject to the given constraint. 10) 10) f(x, y) -4x+ 6y, x2 y2-13 A) Maximum: 36 at (3, 4); minimum: 0 at (0,0) B) Maximum: 26 at (2, 3); minimum: 0 at (0,0) C) Maximum: 36 at (3, 4) minimum: -36 at (3,-4) D) Maximum: 26 at (2, 3); minimum: -26 at (-2, -3)

Explanation / Answer

1.Let us first find the critical points in this function

In order to achieve this goal, we need to find derivatives of the function

f(x,y) = 36x2+18xy+9y2

fx  =df/dx= 72x +18y--------1

fy =df/dy = 18x +18y----------2

fxy = 18

fxx =d2f/dx2 = 18

fyy = d2f/dy2 = 18

for critical points, fx =0 and fy = 0

solving equation 1 and 2,we can see that the only critical point we have is (0,0).

In order to categorize the critical point , we need to define

D(x,y) = fxxfyy-[fxy]2

If D >0 and fxx > 0, the point is local minimum

If D >0 and fxx < 0, the point is local maximum

if D< 0, then point is saddle point

But in our case :

We can see that D = 0

In this case, point can be wither minimum, maximum or saddle point.

So our test has failed :). But atleast we know that there is only 1 critical point, we have to find the catogory of it.So..

Now we have to go for other method:

f(x,y) = 36x2+18xy+9y2

The function can be written as :f(x,y) = 35x2 +x2 +18xy+9y2

f(x,y) = 35x2 + (x+9y)2

As this function will always be > = 0

f(x,y) > f(0,0) ----- for any x,y

This means, at (0,0), which is the critical point, the f(0,0) will be minimum

Hence, f(0,0) = 0, is local minimum, and the function has NO OTHER POINTS

Option (C) is correct

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