Question

11 Find any I the extreme values (if of the given function f(x, y, 2) = x² + 2y? subject to the constraint x²+y²-2²=1 (find minimum, argue that does not exist ) maximum

Answer 1

o se²+ y = f (oc, y, z) = z) = x² + 2y² subject to se+yu - 2 = 1 1+z² z²zo H ZER se² + y z Hence from f(x, y, z) = (x² + 2 y ² ) (2+gs + y 1+z2+ ye cusing Hence (2²+ y) = 0 f(xy, 2) > 1 because min [f(x, y, z) = 1 Hence and Z=0 = 0 at x = 1 be than 1 Less so that x² + y 2 = 1 + 0-0 = 1 Note that 4 (267, 2) = x + 2y2 cannot because if fcae, y, z) <1 60) é +2y <! cor) x + 2y² <1 + 2 cor x + y2 - 2 <1-y² <1 x² + y² - ² = 1 KI we but know

Now and put x = z = vm+1 and y = get Y-2 - z2 = M+lto-M = 1 constraint is satisfied. 2²+ So Now Mti for y, z) = x + 2y = Taking m arbitraily larges can be arbitrarily large. large flaey, 2) ТМ Hence exist. max [fcx, y, z)] does not Hence at (1,0,0) - Min (f (248, zu Max [f(x, y, z) = does not exist