Maximizing f(x,y) on a Triangular Region | Calc 3 Homework Solution

[e^(i Pi)+1=0]

Homework Statement

Find the maximum value of f(x,y)=xy+1 on the triangular region: x ≥ 0, y ≥ 0, x+y ≤ 1

The Attempt at a Solution

Partial derivatives of f(x,y) with respect to both x and y are positive within the given bounds, so the maxima will be found on the boundary where y=1-x. g(x) = -x²+x+1 and max value is 1.25

 

[verty]

How did you reason that y = 1-x? Why could the maximum not be found where either x = 0 or y = 0?

 

[haruspex]

How did you reason that y = 1-x? Why could the maximum not be found where either x = 0 or y = 0?

For the reason given: since both partial derivatives are positive within the region, the max must be at a place where neither x nor y can be increased without reducing the other.

 

[verty]

For the reason given: since both partial derivatives are positive within the region, the max must be at a place where neither x nor y can be increased without reducing the other.

Can we extend this reasoning to the case where the partial derivatives differ in sign? I think so. This is most interesting.

Regarding the original post, I thought perhaps a leap had been made but I see now that everything was justified.

 

[Ray Vickson]

Can we extend this reasoning to the case where the partial derivatives differ in sign? I think so. This is most interesting.

Regarding the original post, I thought perhaps a leap had been made but I see now that everything was justified.

No, it is not true in general. For example, if we want to maximize f = 2*x + y over the region x,y >= 0, x+y <= 1, the solution is at (x,y) = (1,0). This is true despite the fact that the partial derivatives of f are positive everywhere in the feasible set.

The OP happened to get lucky in his analysis, but other, more general cases are not so easily handled. I, personally, would not regard his argument as 100% airtight, but he could fairly easily fix it up.

The so-called Karush-Kuhn-Tucker conditions were devised to deal with such problems. Essentially, these look at the behavior of directional derivatives of f along feasible directions.

 

[pasmith]

No, it is not true in general. For example, if we want to maximize f = 2*x + y over the region x,y >= 0, x+y <= 1, the solution is at (x,y) = (1,0).

... which lies on the line y = 1 - x.

 

[Ray Vickson]

... which lies on the line y = 1 - x.

Yes, of course it does. However, that does not prevent one of the variables x or y from being zero at the solution.

What I was objecting to was the automatic assumption that both variables had to be > 0, just because the partials are both > 0.

 

[haruspex]

What I was objecting to was the automatic assumption that both variables had to be > 0, just because the partials are both > 0.

No, that was in verty's comment, not in the OP. All e^(i Pi)+1=0 argued was that it must lie on x+y=1.

 

[e^(i Pi)+1=0]

I know the logic doesn't hold when x or y = 0, but that obviously doesn't give the max value just by glancing at the function so I didn't think it was worth mentioning.

 

[haruspex]

I know the logic doesn't hold when x or y = 0

You are selling your own argument short. Even if, say, x = 0, if 0 < y < 1 then, because ∂f/∂x > 0, must be possible to increase the value of the function by increasing y. The case where it does break down is x = y = 0, because both partial derivatives are 0 there.