Find extremum of function $z(x,y) = 2x^2-y^2-6x$ under condition $A(x,y) = x^2+y^2 \le 9$. How can we manage when the condition is given by inequality? I only know that Lagrange multiplier works with condition given by equality.
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1You can introduce a slack variable $w\ge 0$, which changes the condition to $x^2+y^2+w=9$ – John Doe Jan 12 '18 at 15:05
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1You could first find all interior extrema, where the constraint is not binding. And then all extrema on the boundary, that is, where the constraint is $x^2+y^2=9$. – Gerhard S. Jan 12 '18 at 15:07
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let $$f(x,y)=2x^2-y^2-6x$$ the you will get $$\frac{\partial f(x,y)}{\partial x}=4x-6$$ and $$\frac{\partial f(x,y)}{\partial y}=-2y$$ we get $$x=\frac{3}{2},y=0$$ and then you Can calculate the extrema on the curve $x^2+y^2=9$ so you will have $$f(x,\pm\sqrt{9-x^2})=2x^2-(9-x^2)-6x$$ Can you finish? $$f'(x,\pm\sqrt{9-x^2})=6x-6$$
Dr. Sonnhard Graubner
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This method is also missing the extremum $(x,y)=(-3,0)$, the constrained maximum, because of the parametrization $y=\sqrt{9-x^2}$. Now I see this problem is full of traps. – Zhuoran He Jan 12 '18 at 15:22
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For $x=-3$ and $y=0$ we obtain a value $36$.
We'll prove that it's a maximal value.
Indeed, we need to prove that $$2x^2-y^2-6x\leq36$$ and since $$36\geq3(x^2+y^2)+9,$$ it's enough to prove that $$2x^2-y^2-6x\leq3(x^2+y^2)+9$$ or $4(x+3)^2+4y^2\geq0,$$ which is obvious.
For $x=1$ and $y=2\sqrt2$ we get a value $-12$.
We'll prove that it's a minimal value.
Indeed, we need to prove that $$2x^2-y^2-6x\geq-12$$ and since $$12\geq3+x^2+y^2,$$ it's enough to prove that $$2x^2-y^2-6x+3+x^2+y^2\geq0$$ or$$3(x-1)^2\geq0.$$ Done!
Michael Rozenberg
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