Find Minimum value of $$f(x)=\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}$$
My try: the domain of the function is $x \in [-1 \,\,\,1]$
Differentiating and equating it to zero we get
$$f'(x)=\frac{-21}{\sqrt{58-42x}}+\frac{70}{\sqrt{1-x^2}\sqrt{149-140\sqrt{1-x^2}}}=0$$
but its very tedious to find critical points here.
any other approach?
Best Answer
Let $(x,y)$ be a point on the unit circle $x^2+y^2=1$. We have to minimize the function: $$ \begin{aligned} f(x,y) &= \sqrt{(7x-3)^2+(7y-0)^2} \ +\ \sqrt{(7x-0)^2+(7y-10)^2} \\ &= \operatorname{Distance}(\ (7x,7y)\ ,\ (3,0)\ )\ +\ \operatorname{Distance}(\ (7x,7y)\ ,\ (0,10)\ ) \\ &\ge \operatorname{Distance}(\ (3,0)\ ,\ (0,10)\ ) =\sqrt{3^2+10^2}\ , \end{aligned} $$ with equality in the $\ge$ above only for the point of intersection of the segment with the above distance with the circle of radius $7$ centered in the origin.