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Let $c>0$ be a given constant and consider the real function on $\mathbb R^2$ given by $f(x,y)=c(x^2+y^2)$. Is there an easy way (that is, a solution without using Lagrange multipliers) to determine a $c$ such that that minimal value of $R(x,y,z)=x^2+y^2+(z-5)^2$ on the surface given by the equation $f(x,y)=z$ is less than or equal to $1$?

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Have I interpreted something wrong here? Based on what you've written, $f(x,y) = 0 \Leftrightarrow x^2+y^2 = 0$, and so on this surface we have $R(x,y,z)=(z-5)^2$, which has minimum value $0$ at $z=5$, for any value of $c$. – Clive Newstead Jul 30 '12 at 12:46
Perhaps you meant $z=f(x,y)$? – joriki Jul 30 '12 at 13:06
The title and body of the question don't match; please make the title accurately summarize the question. – joriki Jul 30 '12 at 13:19

If you mean $f(x,y)=z$ rather than $f(x,y)=0$ then since $x^2+y^2 = \dfrac{f(x,y)}{c}$, we obtain that on the surface $f(x,y)=z$ we have

$$G(x,y,z) = \dfrac{z}{c} + (z-5)^2$$

This is just a quadratic polynomial, whose minimum value can be found by completing the square: doing so, we get

$$G(x,y,z) = \left( z + \dfrac{1}{2c} - 5 \right)^2 + 25 - \left( \dfrac{1}{2c} - 5 \right)^2$$

This has minimum value $25 - \left( \dfrac{1}{2c} - 5 \right)^2$ $(*)$, and so we require $$\left( \dfrac{1}{2c} - 5 \right)^2 > 24$$ i.e. $$\boxed{c < \dfrac{1}{10+4\sqrt{6}}}$$

Much of the above is unnecessary, since we necessarily have $z \ge 0$ and so the minimum value $(*)$ might not be attained. But it means that you certainly don't have to use any calculus at all.

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Or any $c\gt\frac52+\sqrt6$, for example $c=5$. – Did Jul 30 '12 at 14:03

If you mean the surface $z=f(x,y)$, then consider the point $(1/2,1/2,c/2)$. The value of $R$ at this point is $$ \frac{1}{2} + \left(\frac{c}{2}-5\right)^2 $$ Setting $c=10$ makes this $1/2$. Hence the minimum is $\leq 1/2 < 1$

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