How Do You Find the Absolute Min and Max of a Function on a Triangular Region?

[riri]

Hello :)

I'm struggling a bit on how I can find steps to solve problems like this, because my mind goes all over the place.
Find absolute min and max of f(x,y) = 18-3x+6y on a closed triangular region with vertices (0,0), (6,0), and (6,9).

I still struggle with partial derivatives and take a lot of time to find them and if anyone knows simple steps I can use to solve problems like this that would be so helpful.

For now, I drew a picture with the 3 vertices, and I used pythagorean theorem to get the last line which gave me 10.816 (I don't think this is relevant though).

I found the partial derivatives -3 and 6.
The objective function is 18-3x+6y but I'm stuck on what the constraint is. Is it 3x-6y=-18?

Then I use lagrange which gives me -1.

Anyway, I plugged in random things and got to what is supposedly the right answer of max=54 and min=0. I'm just having hard time finding the way to get to the answer for these types of questions.
Any help is appreciated! :)

 

[Theia]

For these kind of problems, does it help if we list what possibilities there are? I mean, you can find a minimum or a maximum at a point where 1) partial derivatives are equal to 0 or 2) on the bound of the region.

In this case, you already calculated the derivatives and clearly they can't vanish at any point. So... What you can say about the value of the function on the bound of the region?

 

[riri]

Thank you! Okay, I understand the partial derivates can't vanish part :)
But for the bounded region,
So I would just plug in the points given by the vertices into the function right?

 

[Theia]

But for the bounded region,
So I would just plug in the points given by the vertices into the function right?

Basically the answer depends on what kind of answer you are looking for. :D

Case I: In linear optimization problems - like this problem - it is enough to check the value of the function on the vertices. So yes, you are right. But it is important to note that this is not enough when the function is not linear, but e.g. $$f(x,y) = x^2y$$.

Case II: In general, you should find the functional shape of the boundary line (say that is $$y = g(x)$$ for some part of the boundary line), plug in the function $$f = f(x,g(x))$$ and analyze if there is an extremum.

Since the function $$f$$ is now linear, and the boundary line is also a (part of) linear function, the resulting function $$f = f(x,g(x))$$, which you should analyze, is still linear. And being such, the extrema are found by inserting the vertices into the function in question.

 

[HallsofIvy]

Hello :)

I'm struggling a bit on how I can find steps to solve problems like this, because my mind goes all over the place.
Find absolute min and max of f(x,y) = 18-3x+6y on a closed triangular region with vertices (0,0), (6,0), and (6,9).

I still struggle with partial derivatives and take a lot of time to find them and if anyone knows simple steps I can use to solve problems like this that would be so helpful.

For now, I drew a picture with the 3 vertices, and I used pythagorean theorem to get the last line which gave me 10.816 (I don't think this is relevant though).

That is the length of the hypotenuse which doesn't help. What you need to know is the equation of the line. The line through (0, 0) and (6, 9) has equation y= (9/6)x= (3/2)x.

I found the partial derivatives -3 and 6.
The objective function is 18-3x+6y but I'm stuck on what the constraint is. Is it 3x-6y=-18?

Then I use lagrange which gives me -1.

Anyway, I plugged in random things and got to what is supposedly the right answer of max=54 and min=0. I'm just having hard time finding the way to get to the answer for these types of questions.
Any help is appreciated! :)

The fact that the partial derivatives are not 0 in the interior of the triangle means there is no local max or min there so there can be no absolute max or min in the interior.

You need to check the boundaries now. The line through (0, 0) and (6, 0) has equation y= 0. Setting y= 0 in 18- 3x+ 6y gives 18- 3x. Its derivative is the non-zero constant -3 so there is no max or min on that line. The line through (6, 0) and (6, 9) has equation x= 6. Setting x= 6 in the function gives 18- 18+ 6y= 6y. Its derivative is the non-zero constant 6 so there is no max or min on that line. The hypotenuse has equation y= (3/2)x as I said above. Putting that in the function gives 18- 3x+ 6(3/2)x= 18- 3x+ 9x= 18+ 6x. Its derivative is the non-zero constant 6 so there is no max or min on that line. Now we need to check the boundaries of those lines- the points (0, 0), (6, 0) and (6, 9). Set those in the function 18- 3x+ 6y. The maximum of the function on that set is the largest of those three values, the minimum is the smallest.