Find the absolute maximum and minimum values of f on the set D.?

To find global extrema identify all feasible unconstrained extrema of f. Then identify any potential extrema along the boundary. Out of the list generated select the greatest and the least. Since the global extrema are contained in this set you don’t need to categorize individual points as local max or min..

First find unconstrained CP by solving

fₓ = 4−2x = 0 ⟹ x=2

fᵧ = 6−2y = 0 ⟹ y=3

f(2,3)=15 and (2,3) is feasible and so is a candidate for global extrema

Now find extreme values along boundary in 4 sections. You can do corners separately to avoid repeating end-point calculations..

f(0,0)=2, f(0,5)=7, f(4,5)=7, f(4,0)=2

x=0, 0<y<5 : f(0,y)=6y−y²+2 which has a CP at y=3 which is feasible and f(0,3)=11 .

x=4, 0<y<5 : f(4,y)=6y−y²+2 which has a CP at y=3 which is feasible and f(4,3)=11

y=0, 0<x<4 : f(x,0)=4x−x²+2 which has a CP at x=2 which is feasible and f(2,0)=6

y=5, 0<x<4 : f(x,5)=4x−x²+7 which has a CP at x=2 which is feasible and f(2,5)=7

Global max is f=15 at (2,3)

Global min is f=2 at (0,0), (4,0)

f(x,y) = 4x+6y-x^2-y^2+2

∂f/∂x = 4 -2x = 0

2x=4

x=2

∂f/∂y = 6-2y=0

2y=6

y=3

∂^2f/∂x^2 = -2

∂^2f/∂y^2 = -2

∂^2f/∂x∂y = 0

Maximum and minimum for functions of 2 variables

Fin∂ D(x,y)=(∂^2f/∂x^2) (∂^2f/∂y^2) - (∂^2f/∂x∂y)^2

If D(a,b) > 0 an∂ ∂^2f/∂x^2 > 0, then f(x,y) has a minimum at (a,b)

If D(a,b) > 0 an∂ ∂^2f/∂x^2 < 0, then f(x,y) has a maximum at (a,b)

If D(a,b) < 0 an∂ , then f(x,y) has neither a maximum nor a minimum at (a,b)

If D(a,b) = 0 an∂ , then no conclusion can be ∂rawn from this limit

D(2,3) = (-2)(-2) - 0 = 4

D(2,3) > 0 an∂ ∂^2f/∂x^2 < 0, then f(x,y) has a maximum at (2,3)

Absolute maximum = (4)(2)+6(3)- (2)^2 -(3)^2 + 2 = 15

Absolute minimum occurs at (0,0) = 2