What Is the Minimum Value of $f$ with Given Conditions?

[Albert1]

find minimum value of $f$
given :
$(1)a,b,c,x,y\in R^+$
$(2)x+y=c$
$(3)f=\sqrt {a^2+x^2}+\sqrt {b^2+y^2}$
find :$min (f)$

 

[lfdahl]

My solution:

Spoiler

I use raw calculus (Lagrange optimzation) and I would be happy to see other approaches:

\[f(x,y) = \sqrt{a^2+x^2}+\sqrt{b^2+y^2}\]

The constraint: $g(x,y) = x+y -c = 0$

Solving for $x (>0)$:

\[\frac{\partial f}{\partial x}=-\lambda \frac{\partial g}{\partial x} = -\lambda \: \: \: \wedge \: \: \: \frac{\partial f}{\partial y}=-\lambda \frac{\partial g}{\partial y} = -\lambda\]

\[ \Rightarrow \frac{x}{\sqrt{a^2+x^2}} = \frac{y}{\sqrt{b^2+y^2}}=\frac{c-x}{\sqrt{b^2+(c-x)^2}}\]

\[ \Rightarrow x^2(b^2+(c-x)^2)=(c-x)^2(a^2+x^2)\]

\[ \Rightarrow (b^2-a^2)x^2+2a^2cx-a^2c^2=0\]

\[ \Rightarrow x = \frac{ac}{a+b} \Rightarrow y = c-x = \frac{bc}{a+b}\]

- and the minimum value of f is: $f_{min} = \sqrt{(a+b)^2+c^2}$.

 

[Albert1]

find minimum value of $f$
given :
$(1)a,b,c,x,y\in R^+$
$(2)x+y=c$
$(3)f=\sqrt {a^2+x^2}+\sqrt {b^2+y^2}---(1)$
find :$min (f)$

using geometry

Spoiler

consturct $\triangle ACD,\angle A=90^o, \overline {AC}=a+b=\overline {AB}+\overline {BC}$
$\overline {DA}=c=\overline {DQ}+\overline {QA}=x+y$
point $P$ is an inner point of $\triangle ACD,\overline {PQ}=a=\overline {AB},$ and $\overline {PQ}\perp \overline{AD}$
$\overline {PB}=y=\overline {AQ},$ and $\overline {PB}\perp \overline{AC}$
we have $(1):f=\overline {CP}+\overline {PD}\geq\overline {CD}=\sqrt{(a+b)^2+c^2}$
equality holds when point $P $ locates on $\overline {CD}$