+535 The temperature of a point (x,y) in the plane is given by the expression x^2 + y^2 - 4x + 2y - 12x + 14y + 26. What is the temperature of the coldest point in the plane?
+1926 First, let's combine some like terms to simplify this expression!
\(x^2 -14x + y^2 + 16y + 26 \)
Now, all we have to do is to take the derivative with respect to x and y and set that to 0 before plugging the values back into the function.
We have
\(2x - 14 = 0 \\ 2x = 14 \\ x = 7\)
and
\(2y + 16 = 0 \\ 2y = -16 \\ y = -8\)
Now, plugging the value back into the function, we have \((7)^2 - 14(7) + (-8)^2 + 16(-8) + 26 = -87 \).
So the lowest temperature is just -87!
Thanks! :)
