Given the function $f(x,y)=4x+2y$, I wish to find the extreme points (min,max) given the following constraint: $y=x^{2}+1, -3\leq x\leq 0$

The constraint contains both a condition and a region. Which way should I be using here ? Should it be lagrange multipliers, or should it be an extrema under a bounded region ?

In the second case, how do I find the min/max ? Can I say that the edges points over $y=x^{2}+1$ are (0,1) and (3,10) and there are no local points?

Thank you !

  • $\begingroup$ You should draw a sketch of the part of the graph of $(x,y)$ for the range of interest. Then lightly rule a few of the contour lines for $f$. That should make it clear which parts of your problem are solvable by calculus means and which are not. $\endgroup$ – kimchi lover Jul 19 '17 at 12:27

HINT: consider the function $$f(x,y)=f(x,x^2+1)=4x+2(x^2+1)$$ and you got a Problem in only one variable

  • $\begingroup$ Your method give identical results to lagrange multipliers, i.e., one point which is (-1,2) and the value of f there is 0. Am I correct ? $\endgroup$ – user3275222 Jul 19 '17 at 11:09
  • $\begingroup$ yes that is correct, there is a global Minimum. $\endgroup$ – Dr. Sonnhard Graubner Jul 19 '17 at 11:11
  • $\begingroup$ and $$f_x\ne 0$$ and $$f_y\ne0$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 19 '17 at 11:13
  • $\begingroup$ I agree, there are no local min/max. My book said that the sum of value of the extrema points is 8. It's a mistake, right ? $\endgroup$ – user3275222 Jul 19 '17 at 11:14
  • $\begingroup$ can you post the original Problem please? $\endgroup$ – Dr. Sonnhard Graubner Jul 19 '17 at 11:17

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