Thread: Find the absolute maximum and minimum of...

Find the absolute maximum and minimum of f(x,y)=x^2+xy+y^2-6y on the rectangle {f(x,y) | -3 <= x <=3, 0 <= y <= 5}

i find the critical point by taking the derivative of f(x,y), setting it to 0, then finding x and y.

fx = 2x+y
fy= x+2y-6

y = -2x

x + 2(-2x) - 6 = 0
x - 4x = 6
-3x = 6
x = -2

y = -2(-2)
y = 4

i'm unsure of what to do next, can someone help? thanks!

Now evaluate the Hessian matrix at that point to determine the nature of the stationary point.

okay so

Hf(-2,4) = [2,1]
[1,2]

d1 = fxx = 2
d1 = 2 > 0

d2 = (2)(2) - (1)(1) = 3
d2 = 3 > 0

local minimum = (-2,4)

did i do this correctly? how do i find the maximum? is local and absolute the same?

i think i got it.

since local minimum is (-2,4) then
f(-2,4) = (-2)^2 + (-2)(4) + (4)^2 - (6)(4) = -12
so
Absolute Minimum = -12

and for absolute max you take the highest point within the range of {f(x,y) | -3 <= x <=3, 0 <= y <= 5}
so, (3,5)
and
f(3,5) = (3)^2 + (3)(5) + (5)^2 - (6)(5) = 19
so
Absolute Minimum = 19


is this right? is the way i found the point to solve the absolute maximum the right way to do it?

Well, what did you do find the maximum? Just evaluate at the point with largest x and y? There is no reason to think that will give a maximum. Nor is there reason to think that a local minimum has to be an absolute minimum. There is really no need to determine the Hessian since you are looking for absolute max and min and finding out if it is a local max or min is irrelevant.

The theorem says that max and min of a differentiable function occur where the derivative is 0 or on the boundary. Once you have found that (-2, 4) is the only point in the interior where the derivative is 0, turn to the boundary- which, in this case, consists of the four lines, x= -3, x= 3, y= 0, and y= 5.

On y= 0, $\displaystyle f(x,0)= x^2$ which has derivative 2x so its derivative is 0 at (0, 0).
On y= 5 $\displaystyle f(x, 5)= x^2+ 5x+ 25-30= x^2+ 5x- 5$. Its derivative is 2x+ 5 which is 0 when x= -5/2. Another critical point is at (-5/2, 0).
On x= -3, $\displaystyle f(-3, y)= 9- 3y+ y^2- 6y= y^2- 9y+ 9$. Its derivative is 2y- 9 which is 0 when y= 9/2. Another critical point is at (0, 9/2).
On x= 3, $\displaystyle f(3, y)= 9+ 3y+ y^2- 6y= y^2- 3y+ 9$. Its derivative is 2y- 3 which is 0 when y= 3/2. Another critical point is at (0, 3/2).

And, of course, don't forget the endpoints of those intervals, the vertices of the rectangle, (-3,0), (-3, 5), (3, 0), and (3, 5).

Evaluate the function, $\displaystyle f(x, y)= x^2+ xy+ y^2- 6y$ at those points:
(-2, 4), (0, 0), (-5/2, 0), (0, 9/2), (0, 3/2), (-3, 0), (-3, 5), (3, 0), and (3, 5). The largest value you get for those points is the maximum and the smallest is the minimum on the entire rectangle.