I have to find absolute maximum and minimum values of $f(x,y)$ = $4x^{2} + 9y^{2} -8x - 12y + 4 $

over rectangle in first quadrant bounded by lines $x=2 , y=3$ and coordinate axes

I have checked interior points for maxima and minima .But for points on the boundary i need to check now .I can also substitute values in original function of two variables and reduce it into single variable and check extremum on boundary . But im interested to use lagrange multipliers in this case . Can anyone help me with that ? Thanks in advance

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    $\begingroup$ Do every side in turn: first the function subject to $\;x=0\;$ , then subject to $\;x=2\;$ , then to $\; y=0\;$ and finally to $\;y=3\;$ . Unless this is a rather annoying exercise, I can't understand why would you want Lagrange multipliers instead of substitution. $\endgroup$ – Timbuc Nov 13 '14 at 11:55
  • $\begingroup$ @Timbuc thanks . i was trying constraint equation $xy=6$ for this . $\endgroup$ – Sophie Clad Nov 13 '14 at 11:57
  • $\begingroup$ Why would you do that, @Sophie ? That hyperbola touches your rectangle in only one point, namely $\;(2,3)\;$ . $\endgroup$ – Timbuc Nov 13 '14 at 12:00
  • $\begingroup$ @Timbuc oh! yes thanks . $\endgroup$ – Sophie Clad Nov 13 '14 at 12:03

Lagrange multipliers seems overkill.$$f(x, y) = 4x^2 + 9y^2 -8x - 12y + 4 = 4(x-1)^2+(3y-2)^2-4$$

Clearly it gets a global minimum of $-4$ when $x = 1, y = \frac23$ which is within the feasible region. Further, as it is convex, the maximum has to be at the boundary corners, so we need only to check $f(0, 0), f(2, 0), f(0, 3), f(2, 3)$ to find the maximum at $f(0, 3)=49$.

  • $\begingroup$ What is meant by convex region , and so extremum has to be at corners ?? $\endgroup$ – godonichia Nov 13 '14 at 13:28
  • $\begingroup$ @godonichia A convex function e.g. $f(x, y)$, defined on a closed convex polygon (any line segment connecting points in the polygon, itself lies in the polygon), will have maxima only at its corners... $\endgroup$ – Macavity Nov 13 '14 at 13:35
  • $\begingroup$ ok as i have understood any region enclosed by joining line segments (not curves) eg square ,triangle , rectangle , in general polygon we only need to check intersection of line segments i.e (corners) . am i correct in gettting ? $\endgroup$ – godonichia Nov 13 '14 at 13:37
  • $\begingroup$ @godonichia Correct. This is because corners are the only points which are not convex combinations of other points in the polygon. $\endgroup$ – Macavity Nov 13 '14 at 14:26

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