Find the absolute maximum and minimum values of

f(x, y) = 2 + 2x + 2y – x2 – y2

on triangular plate in the first quadrant, bounded by the lines x = 0, y = 0 and y = 9 – x.

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UPSC ESE 2020 Paper 1
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  1. -4
  2. -2
  3. 4
  4. 2

Answer (Detailed Solution Below)

Option 3 : 4
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Detailed Solution

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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.

F1  RaviRanjan 18-01-22 SavitaD1

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
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