# Show that Maximum and Minimum are Global

So I have the following question:

Find the extrema of the function $$f(x,y)=4x-6y$$ Given the constraint $$4x^2-4x+9y^2-6y-2=0$$ And determine whether these extrema are local/global on the constraint.

I found a max and min respectively at $$(\frac{1+\sqrt2}{2},\frac{1-\sqrt2}{3}) ,(\frac{1-\sqrt2}{2},\frac{1+\sqrt2}{3})$$ with the value of f(x,y) at those points being $$4\sqrt2 ,-4\sqrt2$$

I know these points a global maxima on the restriction/constraint, but I am having trouble proving that they are global.

• Show that for when $4x-6y=4\sqrt{2}$ there is only one solution of the constraint equation and when $4x-6y>4\sqrt{2}$ there is no solution. This can be done by substituting for $x$ or $y$ in the constraint equation and examining the resulting quadratic equation. Do the corresponding process at the other end, and you are done. May 11, 2020 at 0:18
• @Peter, I did that, and found a solution for the > part. I am not really sure what you are getting at. May 11, 2020 at 0:35
• If the max value of $4x-6y$ is $4\sqrt{2}$, then there is no solution for $(x,y)$ if $4x-6y>4\sqrt{2}$. If the min value of $4x-6y$ is $-4\sqrt{2}$, then there is no solution if $4x-6y<-4\sqrt{2}$. May 11, 2020 at 1:00

Your constraint is an ellipsoid and your objective function is a straight line on 2-D. Therefore the straight line tangent to the ellipsoid should give you the global maximum/minimum.

weimiao gives the essential answer to your question: when the function is linear, changing the value of $$\ ax \ + \ by \ = \ c \$$ shifts the line graph of the function parallel to itself. For a closed bounded constraint curve, there will be tangent lines to that curve which have largest and smallest values of $$\ c \$$ . There is something of further interest to remark about this situation.

When we have a constraint curve which has four-fold symmetry about the origin, a linear function will have level curves $$\ ax \ + \ by \ = \ \pm c \$$ at tangent points that are symmetric about the origin $$\ ( x_m \ , \ y_m ) \$$ and $$\ ( -x_m \ , \ -y_m ) \$$ . This symmetry can be often exploited to find the absolute maximum $$\ +c \$$ and minimum $$- c \$$ for the function fairly easily; this can also be the case for other sorts of functions.

What is interesting about this extremization problem is that the constraint curve $$4x^2 \ - \ 4x \ + \ 9y^2 \ - \ 6y \ \ - \ \ 2 \ \ = \ 0 \ \ \rightarrow \ \ \frac{(x \ - \ \frac{1}{2})^2}{1} \ + \ \frac{(y \ - \ \frac{1}{3})^2}{4/9} \ \ \ = \ \ 1 \ \ ,$$

while an ellipse with its major and minor axes parallel to the coordinate axes, is not centered on the origin, and thus has no symmetry about the origin. We expect that the tangent points to this ellipse for parallel lines are not symmetrically placed in this way, so the maxima and minima of linear functions will not generally be the negatives of one another.

The exception happens to be the linear function $$\ 4x \ - \ 6y \$$ . For the value $$\ c \ = \ 0 \$$ , the line $$\ 4x \ - \ 6y \ = \ 0 \$$ passes through the origin and the center of the constraint ellipse, so the ellipse is symmetrical about that line. We could then choose coordinate variables relative to the center of the ellipse, $$\ X \ = \ x \ - \ \frac{1}{2} \ \ , \ \ Y \ = \ y \ - \ \frac{1}{3} \$$ to write the constraint curve as $$\ \ 4X^2 \ + \ 9Y^2 \ = \ 4 \ \$$ and the linear function as

$$\ 4 · \left(x \ - \ \frac{1}{2} \right) \ - \ 6 · \left(y \ - \ \frac{1}{3} \right) \ = \ c \ - \ 2 \ + \ 2 \ \ \rightarrow \ \ 4X \ - \ 6Y \ = \ c \ \ .$$

We can now solve for the absolute extrema of the linear function without a sophisticated technique such as Lagrange multipliers. Implicit differentiation of the constraint curve yields

$$\ 8X \ + \ 18YY' \ = \ 0 \ \ \Rightarrow \ \ Y' \ = \ \frac{-8X}{18Y} \ = \ -\frac{4X}{9Y} \ \ .$$

Since the linear function produces parallel lines $$\ Y \ = \ \frac{2X - c}{3} \ , \$$ we find

$$Y' \ = \ -\frac{4X}{9Y} \ \ = \ \ \frac{2}{3} \ \ \Rightarrow \ \ -12X \ = \ 18Y \ \ \Rightarrow \ \ Y \ = \ -\frac{2}{3}X$$

(for instance) $$\Rightarrow \ \ 4X^2 \ + \ 9 · \left(-\frac{2}{3}X \right)^2 \ \ = \ \ 4 \ \ \Rightarrow \ \ 4X^2 \ + \ 4 X^2 \ \ = \ \ 4 \ \ \Rightarrow \ \ X^2 \ = \ \frac{1}{2}$$ $$\Rightarrow \ \ X \ = \ \pm \frac{\sqrt{2}}{2} \ \ , \ \ Y \ = \ -\frac{2}{3} \ · \ \pm \frac{\sqrt{2}}{2} \ \ = \ \ \mp \frac{\sqrt{2}}{3} \ \ .$$

The absolute extrema for our linear function are

$$f(X,Y) \ \ = \ \ 4X \ - \ 6Y \ \ = \ \ 4· \left(\pm \frac{\sqrt{2}}{2} \right) - \ 6· \left(\mp \frac{\sqrt{2}}{3} \right) \ = \ \ \pm 2 \sqrt{2} \ \pm \ 2 \sqrt{2} \ \ = \ \ \pm 4 \sqrt{2} \ \ .$$

So this symmetry occurs even though the associated tangent points, which you show, are not symmetrical about the origin:

$$X \ = \ \frac{\sqrt{2}}{2} \ , \ Y \ = \ -\frac{\sqrt{2}}{3} \ \ \rightarrow \ \ \left( \frac{1 + \sqrt{2}}{2} \ , \ \frac{1 - \sqrt{2}}{3} \right) \ \ \approx \ \ (1.207 \ , \ -0.138) \ \ \text{for} \ \ f \ = \ 4\sqrt{2} \ \ ,$$ $$X \ = \ -\frac{\sqrt{2}}{2} \ , \ Y \ = \ \frac{\sqrt{2}}{3} \ \ \rightarrow \ \ \left( \frac{1 - \sqrt{2}}{2} \ , \ \frac{1 + \sqrt{2}}{3} \right) \ \ \approx \ \ (-0.207 \ , \ 0.805) \ \ \text{for} \ \ f \ = \ -4\sqrt{2} \ \ ,$$