Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ vertices

Question

Can't find absolute minimums and maximums!

Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ on the set D where D is the closed triangular region with vertices $(8,0), (0,6)$, and $(0,-6)$.

To find critical points:

$F_x = 0$

$F_y = 0$

$F_x$ (first order derivative with respect to x) $ = 2x - 4$

$F_y = 2y$

Setting them equal to $0$, to find critical points:

$2x - 4 = 0$

$2y = 0$

critical point: $(2,0)$

...

Now, the boundary of the triangle can be expressed in 3 lines:

Left side of triangle, $x = 0$:

Function can be expressed by the one variable function

$f(0, y) = y^2 + 9$

Absolute maximums:

You evaluate those at the given points, and I think also supposed to evaluate at the critical point? (derivative equal to $0$ etc)

$(0,6), (0,-6)$

Absolute minimums:

??? (8,0) <- why not at these points? If y = 0, then that's when it has the absolute minimum, no?

...

Upper right side of the triangle:

between points: $(0,6)$ and $(8,0)$

$y= \frac{-3}{4} x + 6 $

Function can be expressed by the one variable function

$f(x, \frac{-3}{4} x + 6 ) = \ (\frac{-3}{4} x + 6)^2 + x^2 - 4x + 9$

Absolute maximums:

$(0, 6)$

Absolute minimums:

(8,0) INCORRECT, WHY???

...

Lower right side of the triangle:

between points: $(0,-6)$ and $(8,0)$

$y = \frac{3}{4} x - 6$

Function can be expressed by the one variable function

(3/4 x - 6)^2 + x^2 - 4x + 9

Absolute maximums:

????

Absolute minimums:

????

p.s. here's the screenshot

Answer 1

I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.

edit 1:(add more details)

$f(x,y)=(x-2)^2+y^2+5 \implies (x-2)^2+y^2=f(x,y)-5=R^2$

you want to find max and min of $f(x,y)$ is same to find max and min of $R$.

note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$

if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.

Answer 2

Complete the square to have $f(x,y)=y^2+(x-2)^2+5.$

Then clearly the minimum is $5$ for the whole plane. This occurs at the point $(2,0).$ This is in the triangular region, so this is the minimum value of $f$ restricted to this domain.

The maximum occurs at one of the corners of the isosceles triangle. Direct computation shows that it occurs at the base corners $(0,6),(0,-6),$ with value $45.$

Answer 3

Chenbi points out that this problem is pretty easy to visualize and check your results against your intuition.

But, you want a systematic approach, or see where your approach seems to be failing.

Check the interior.

Find $F_x, F_y$ and set these partial derivatives equal to $0.$

Looks like you were fine with this.

$2x - 4 = 0\\ 2y = 0$

$(2,0)$ is a critical point.

We should perform second derivative tests to see if this is a max, min or saddle.

$F_{xx} = 2\\F_{yy} = 2\\F_{xy} = 0$

$F_{xy}^2 < 4F_{xx}F_{yy}$ It is not a saddle.

It must be a minimum.

$F(2,0) = 5$

Next, we check the perimeter. You may need to find parametric equations for the boundary, but in this case, lines are easy.

$x = 0$

$f(y) = y^2 + 9\\ f'(y) = 2y = 0$

Has a critical point at $(0,0)$ This is the minimum point on this edge, but it is not the minimum of for the entire region.

Check the endpionts: $F(0,6) = F(0,-6) = 45$

This is the maximum we have found so far.

Move to the next line.

$y = \frac {3}{4} x -6$

$f(x) = (\frac 34 x -6)^2 + x^2 -4x + 9 = \frac {25}{16} x^2 - 13x + 45\\ f'(x) = \frac {25}{8} x - 13 = 0\\ x = \frac {104}{25}$

This is another sort of local minimum. It is the lowest point on the boundary but not the lowest point overall.

Endpoints... we have already checked $(0,-6)$ What about $(8,0)$?

$F(8,0) = 41$ Not as big as $F(0,6), F(0,-6)$

The line $y = -\frac {3}{4} x +6$ will give similar results as the line $y = -\frac {3}{4} x +6$

Max at $(0,6), (0,-6) = 45$

Min at $(2,0) = 5$