Question

Find the maximum and minimum value of f(x,y)=x^2-xy+y^2+4 on the closed area in the first quadrant bounded by the triangle formed by the lines x=0, y=4 and y=x. Help!

Answer 1

algebra :) if im not mistaken

Answer 2

our trial and error "area" is this:

Answer 3

our min and max are gonna be at a vertex of this area; so test 0,0; 0,4; and 4,4

Answer 4

f(0,0) = 4

f(0,4) = (4)^2+4 = 20

f(4,4) = (4)^2-(4)(4)+(4)^2+4 = 20

it appears that anything on the horizontal will produce 20

Answer 5

Actually, its calc 3. I wish it were that simple, but all I've figured out so far is to use partial derivatives and get the local max/min (there's only a local min at (0,0) by this method) but there's another step after that I haven't figured out yet.

Answer 6

partials will get you the gradient; and wherever the gradient produces (0,0) you get a critical right?

Answer 7

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

gf = ( 2x-y , 2y -x) ... for the partials

Answer 8

Yes, I got 0,0 as a critical point that way and its a local minimum. I know I'm supposed to check the boundary and I'm looking at my book's example and I have no clue how they parametrized the boundary lines.

Answer 9

can you scan your books example?

Answer 10

Ah, no. I don't have access to a scanner. On the book's example, they had boundaries 0<=x<=2, 0<=y<=3x. They rewrote the boundaries as parametric equations as follows:

C1: r1(t)=ti t E [0,2]
C2: r2(t)=2i+ti t E [0,6]
C3:r3(t)=ti+3tj t E [0,2]

Answer 11

they got vector notation of lines there if i see it right

Answer 12

Yes, they did. I just can't remember how to do it. I know r = r0 +td. I know how to find vector d, but if vector r0 isn't given, what do you do?

Answer 13

OK, we must check three kinds of points:
1) Relative extrema in the interior of the triangle (points where the gradient is zero)
2) Relative extrema in the boundary of the triangle
3) Corner points.

You already found the only point where the gradient is zero: (0,0) which is a local minimum. We must check 2) and 3).

We need to parametrize each side of the triangle. Lets start with the one that connects (0,0) and (4,4).
We want a linear function\[\gamma_1(t)\]such that\[\gamma_1(0) = (0,0)\]and\[\gamma_1(1) = (4,4).\]That is,\[\gamma_1(t) = (x, y) + t(d_x, d_y)\]and\[\gamma_1(0) = (x,y) = (0,0)\]\[\gamma_1(1) = (x,y) + (d_x, d_y) = (0,0) + (d_x, d_y) = (d_x, d_y) = (4,4)\]so our function is\[\gamma_1(t) = t(4,4).\]For the next parts, we do the same thing:\[\gamma_2(0) = (4,4)\]\[\gamma_2(1) = (0, 4)\]\[\gamma_2(t) = (x,y) + t(d_x,d_y)\]\[\gamma_2(0) = (x,y) = (4,4)\]\[\gamma_2(1) = (x, y) + (d_x, d_y) = (4,4) + (d_x, d_y) = (0, 4) \Rightarrow (d_x, d_y) = (-4, 0)\]therefore,\[\gamma_2(t) = (4,4) + t(-4, 0).\]

\[\gamma_3(0) = (0, 4)\]\[\gamma_3(1) = (0,0)\]\[\gamma(t) = (x, y) + t(d_x, d_y)\]\[\gamma_3(0) = (x, y) = (0, 4)\]\[\gamma_3(1) = (x, y) + (d_x, d_y) = (0, 4) + (d_x, d_y) = (0,0) \Rightarrow (d_x, d_y) = (0, -4).\]therefore,\[\gamma_3(t) = (0, 4) + t(0, -4).\]

Now we will find the relative extrema in the boundary of the triangle.\[f(\gamma_1(t)) = f(4t,4t) = 16t^2+4, \quad t \in [0, 1]\]which is now a function of only one variable. We find the extrema the usual way:\[\frac{d}{dt}f(4t,4t) = \frac{d}{dt}\left(16t^2+4\right) = 32t = 0 \Rightarrow t = 0\]so the relative extrema is\[\gamma_1(0) = (0,0).\]

\[f(\gamma_2(t)) = f((4,4) + t(-4,0)) = 4(4t^2-4t+5), \quad t \in [0, 1]\]\[\frac{d}{dt}f((4,4) + t(-4,0)) = \frac{d}{dt}\left(4(4t^2-4t+5)\right) = 32t-16 = 0 \Rightarrow t = \frac{1}{2}\]so the relative extrema is\[\gamma_2(1/2) = (2,4).\]

\[f(\gamma_3(t)) = f((0,4) + t(0,-4)) = 4+(4-4t)^2, \quad t \in [0, 1]\]\[\frac{d}{dt}f((0,4) + t(0,-4)) = \frac{d}{dt}\left(4+(4-4t)^2\right) = 32t-32 = 0 \Rightarrow t = 1\]so the relative extrema is\[\gamma_3(1) = (0,0).\]

Now we simply compare all these points:
\[f(0,0) = 4\]\[f(4,4) = 20\]\[f(2,4) = 16\]\[f(0,4) = 20\]so the minimum is at (0,0) and the maximum at (4,4) and (0,4).