Find the absolute maximum and minimum values of the following function on the given set r

Concept:

Let f(x , y) have continuous second-order partial derivatives in some disc centred at a critical point (a, b), and let

\(D = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left( {{f_{xy}}\left( {a,b} \right)} \right)^2}\)

• If D > 0 and fxx(a, b) > 0, then f has a relative minimum at (a, b)
• If D > 0 and fxx(a, b) < 0, then f has a relative maximum at (a, b)
• If D < 0, then f has a saddle point at (a, b)
• If D = 0, then no conclusion can be drawn

Calculation:

f(x, y) = 2 + 2x + 2y – x2 – y2

fx = 2 – 2x, fy = 2 – 2y

fxx = -2, fyy = -2

fxy = 0

fx = 0 2 – 2x = 0 x = 1

fy = 0 2 – 2y = 0 y = 1

Now, the critical point (a, b) = (1, 1)

At (a, b) = (1, 1)

fxx = -2, fyy = -2, fxy = 0

D = 4 > 0 and fxy < 0

So, the function f(x, y) has a relative maximum at the point (1, 1).

f(x, y) at a point (a, b) = (1, 1) is,

f(1, 1) = 4

The given region is triangular plate in the first quadrant, bounded by the lines x = 0, y = 0 and y = 9 – x. It is represented as shown below.

From the above figure, the critical points are: (0, 0), (0, 9), (9, 0)

Now let us check for the absolute maximum and minimum of f(x, y) at all the critical points.

At (0, 0), f(0, 0) = 2

At (0, 9), f(0, 9) = -61

At (9, 0), f(9, 0) = -61

From the above values,

The absolute minimum value of f(x, y) = -61

The absolute maximum value of f(x, y) = 4

Last updated on Sep 22, 2022

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Video Transcript

given function X Y equals two, X squared plus y squared minus eight Y plus 16. And article two xy says that X squared plus y squared less than 25. Then arrow will be a circle at center +00 and radius site. And it's out of point will be 050 minus 550 minus 50. Now a pop X Y equals X squared plus y squared minus eight Y plus 16. So a proper X apex. That is why do X equal to Dubai do X of X squared plus Y squared minus eight plus 16 that equal to two X. And if the vehicle to do a better way equal to do by the way of X squared plus y squared minus eight Y plus 16 that is two Y minus eight. Now to find critical points we have to do a P X equal to zero and zero. By doing that we can get X equal to zero and y equal to four. So critical point X equal to zero. For now to find local maximum or minimum we have to find a xx that is door to door by door to door next to. That is doable do except two weeks, that is two and Y Y. That is due to a bad door. Why to that is doable doorway of do Y minus eight that is two and X X. Y. That is due to do X. Y. That is Dubai do except doable. That is doable do except two Y minus seven. Article 20. Now X X into Y y minus X square Hold square. At a critical 200.4 is two into two minus particle. Two into two minus zero equal to four. That is A positive. So we have a look. We have a minimum at 04. So zero for X. Y has a local minima. And the minimum value is X. Y. At 0400 squared plus four squared minus 18 to 4 Plus 16. That is zero. So if XY has a minimum 8040. Now we don't have any maximum points. That means the maximum values must be at the age of the art. Uh That is is of the circle X squared plus Y squared equal to five. So the values X. Xy +850 is +50 square plus zero square mm hmm plus 16 minus 18 to 0 equal to 41. And the X Y at minus 50 is 41. Also apex Y. At 05 is one. And the apex Y 80 minus five is 81. Which has greater value than others. So it has a it has a maximum value. So the global maximum is it? Do you want 805? Thank you.

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