+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?
+1952 First, let's simplify the expression by combining like terms.
x2−14x+y2+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=02x=14x=7
and
2y+16=02y=−16y=−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! :)
