The sum of the lengths of the twelve edges of a rectangular box is $140$, and the distance from one corner of the box to the farthest corner is $21$. What is the total surface area of the box?
To find the surface area of the box, I need to know the dimensions of the rectangular faces. Let me label the sides of the box as $a, b, c$, where $a, b, c$ are positive real numbers. Then, the sum of the lengths of the edges is $4(a + b + c)$, and the distance from one corner to the farthest corner is $\sqrt{a^2 + b^2 + c^2}$. Using the given information, I can write two equations: $4(a + b + c) = 140$ and $\sqrt{a^2 + b^2 + c^2} = 21$. I want to solve for $a, b, c$, but these equations look hard to work with. Maybe I can simplify them a bit. The first equation implies that $a + b + c = 35$. The second equation implies that $a^2 + b^2 + c^2 = 441$. If I square the first equation, I get $(a + b + c)^2 = 1225$. Expanding this, I get $a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1225$. But I already know that $a^2 + b^2 + c^2 = 441$, so I can subtract that from both sides and get $2(ab + bc + ca) = 784$. Dividing by $2$, I get $ab + bc + ca = 392$. I recall that the surface area of a box is the sum of the areas of the six faces, which are $2(ab + bc + ca)$. So, the surface area is just twice what I found, which is $2 \times 392 = 784$. 