A flat circular plate has the shape of the region D x y x2

A flat circular plate has the shape of the region D = {(x, y) |x^2 + y^2 Inequality 1}. The plate is heated and the temperature T(x, y) at each point (x, y) Epsilon D is given by T(x, y) = x^2 + 2y^2 - x. Determine the hottest and coldest temperatures on the plate.

Solution

We know that, for T (x), x = a is a minimum if T \' (a) = 0 and T \'\' (a) > 0; Also x = a is a maximum if T \' (a) = 0 and T \'\' (a) < 0 . Here T (x,y) = x² + 2y² - x such that x² + y² 1. Then T/x = 2x -1. ²T /x² = 2 , T/y = 4y,   ²T/y²  = 4 . Thus T/x = 0   when x = 1/2. Also, T/y = 0 when y = 0. Further, at the stationary point , we have T/x =  T/y = 0. Then, the only stationary point is ( 1/2, 0). Since ²T /x² and  ²T/y² are both positive and since there is only one stationary point i.e. (1/2, 0), there is a minimum at ( 1/2 , 0). Also at this point, x² + y² = 1/4 + 0 = 1/4 which is < 1. There is no maximumum. Thus, the coldest point on the plate is ( 1/2, 0) . There is no hottest point.