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 $1$?
Finding an easy way to detemine the minimal distance
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calculus
multivariable-calculus
optimization
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3Have 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 $c$. – 2012-07-30
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0Perhaps you meant $z=f(x,y)$? – 2012-07-30
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0The title and body of the question don't match; please make the title accurately summarize the question. – 2012-07-30