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.

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 f_xx(a, b) > 0, then f has a relative minimum at (a, b)
• If D > 0 and f_xx(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

f_x = 2 – 2x, f_y = 2 – 2y

f_xx = -2, f_yy = -2

f_xy = 0

f_x = 0 2 – 2x = 0 x = 1

f_y = 0 2 – 2y = 0 y = 1

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

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

f_xx = -2, f_yy = -2, f_xy = 0

D = 4 > 0 and f_xy < 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