Suppose a weight is to be held $10 ft$ below a horizontal line $AB$ by a wire in the shape of a $Y$. If the points $A$ and $B$ are $8 ft$ apart, what is the shortest total length of wire that can be used?
I am not sure how the diagram would look like.
And I will probably need help after the diagram is drawn.
Sorry, I did not know how I should name my title.
Thank you.
On
Between three arbitrary vertices of a triangle a Fermat point minimizes the total distances from one point and this $Y-$ node is the Fermat-Torricelli point. The sum of distances is minimum when dynamic equilibrium is established between three equal and (angularly) equi-spaced forces acting on the $Y-$ point node at $120^{0} $ between taut strings.
In the diagram below half the weight hangs either on smooth pulleys at nodes $(A,B)$ or act along taut strings through fixed pegs.
For Y shape, the total wire length is minimal if the three legs meet at $120^\circ$ angles. Hence the two top legs are both 4 ft times $\frac{\sqrt 3}2$ = $2\sqrt 3 $ ft long and the node is half of that below the top line, i.e., the vertical leg is $10-\sqrt 3$ ft long.