# Find absolute maximum and minimum

Find the absolute maximum and absolute minimum of the function $f(x,y)=xy-5y-25x+125$ on the region above $y=x^2$ and on or below $y=27$.

My solution:

Critical point: $(5, 25)$

$f$ has a saddle point at $(5, 25)$ so it can't be used to find absolute min/max (correct me if i'm wrong).

Boundary points: $(-5, 0), (5, 0), (0,27)$

$f$ has absolute max of $250$ at $(-5, 0)$

$f$ has absolute min of $-10$ at $(0, 27)$

These are my answers, but the online system (homework portal) doesn't accept them so they may be wrong. Any ideas?

You have the wrong boundary points. The curves $y=x^2$ and $y=27$ intersect at $(\pm 3\sqrt{3},27)$. The points $(\pm 5,0)$ are not in the region at all.

And since you are finding the extrema of a function of two variables, you need to consider points on the entire boundary of the region, not just the "boundary points." So you also need to look at $(x,27)$ for $-3\sqrt{3} <x <3\sqrt{3}$. You many also need to look at $(x,x^2)$ for the same $x$'s, but since the region is above $y=x^2$ probably not.

It is true that $$\ (5,25) \$$ constitutes a saddle point for the function, and so would not be an extremum over $$\ \mathbb{R} \$$ , but its role changes in the specified parabolic region. Having ascertained that there is no critical point in the interior, we next need to check the boundary.

Along the line $$\ y \ = \ 27 \$$ , the function becomes

$$f(x,y) \ = \ xy \ - \ 5y \ - \ 25x \ + \ 125 \ \longrightarrow \ f(x) \ = \ 27x \ - \ 135 \ - \ 25x \ + \ 125 \ = \ 2x \ - \ 10 \ \ .$$

Since this linear function has no relative extrema, we will evaluate it for now at the endpoints of this portion of the region boundary (noted by Rory Daulton)

$$f(-3 \sqrt{3}) \ = \ 2(-3 \sqrt{3}) \ - \ 10 \ = \ -6 \sqrt{3} \ - \ 10 \ \approx \ -20.39 \ \ ,$$ $$f(+3 \sqrt{3}) \ = \ 6 \sqrt{3} \ - \ 10 \ \approx \ +0.39 \ \ .$$

It isn't entirely clear whether the original problem intended to include the parabolic lower boundary, but we will look at our function on $$\ y \ = \ x^2 \$$ as well. (I suspect it should be included, though: otherwise, the endpoints we just worked with would not be part of the boundary, meaning there would be no absolute extrema for the region, as we shall see.)

On the parabola, the function is

$$\longrightarrow \ f(x) \ = \ x · x^2 \ - \ 5 · x^2 \ - \ 25x \ + \ 125 \ = \ x^3 \ - \ 5 x^2 \ - \ 25x \ + \ 125$$ $$= \ (x - 5)^2 \ · \ ( x+5 ) \ \ .$$

The critical points on this portion of the boundary are found from

$$f'(x) \ = \ 3x^2 \ - \ 10 x \ - \ 25\ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac{10 \ \pm \ \sqrt{400}}{6} \ = \ -\frac{5}{3} \ \ , \ \ 5 ,$$

producing the function values

$$f(5) \ = \ (5 - 5)^2 \ · \ ( 5+5 ) \ = \ 0 \ \ , \ \ f(-\frac{5}{3}) \ = \ (-\frac{20}{3})^2 \ · \ \frac{10}{3} \ = \ \frac{4000}{27} \ \approx \ +148.1 \ \ .$$

So the saddle point at $$\ (5,25) \$$ found in your original analysis is a relative minimum on the parabolic section of the boundary. However, the absolute extrema for the closed parabolic region are

maximum: $$\ f(x,y) \ \ = \ \ \frac{4000}{27} \$$ at $$\ \left( \ -\frac{5}{3} \ , \ \frac{25}{9} \ \right)$$

minimum: $$\ f(x,y) \ \ = \ \ -6 \sqrt{3} \ - \ 10 \$$ at $$\ \left( \ -3 \sqrt{3} \ , \ 27 \ \right)$$