+359 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 simplify the expression by combining like terms.
\(x^2 -14x + y^2 + 16y + 26\).
Ok! Now all we have to do is to take the derivative with respect to x and y and set that to 0!
We have
\(2x - 14 = 0 \\ 2x = 14 \\ x = 7\)
and
\(2y + 16 = 0 \\ 2y = -16 \\ y = -8\)
Now, all we have to do is to plug in these values for x and y into the orginial function.
We get \((7)^2 - 14(7) + (-8)^2 + 16(-8) + 26 = -87\).
The lowest temperature on the plane is -87 degrees, which is quite cold!
Thanks! :)
