Use the triangle inequality.

## Solution

Let O be the location of the cafeteria. Then the total distance traveled by all employees is $\mathrm{S}=10^{*} \mathrm{OA}+20^{*} \mathrm{OB}+30 * \mathrm{OC}=10(\mathrm{OA}+\mathrm{OC})+20(\mathrm{OB}+\mathrm{OC})$. According to the triangle inequality, OA+OC is not less than AC (and equality is achieved if and only if O lies on the segment AC), OB+OC is not less than BC (and equality is achieved if and only if O lies on the segment BC). Therefore, $\mathrm{S}$ is not less than $10 \mathrm{AC}+20 \mathrm{BC}$, and equality is achieved if and only if O coincides with point C. Thus, the optimal location for the cafeteria is office C.

## Answer

in office C.
Three offices A, B, and C of a company are located at the vertices of a triangle. Office A has 10 employees, Office B has 20, and Office C has 30. Where should a cafeteria be built to minimize the total distance traveled by all employees of the company?

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