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:



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


enter image description here

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.

  • $\begingroup$ what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...? $\endgroup$ – Jack Mar 2 '16 at 3:22

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 $45.$


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 $(2,0) = 5$


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