Question

Start this exercise on a new sheet of paper. For every question, you need to present a clear and logical solution, show your computations, and present a clean and legible work. Write your final answer in a box or highlight it. Consider the function 3 f(x, y) 222 +2° + 3y?. 1. Find the local minima, local maxima, and saddle points of f, and the points where they occur. 2. Use the method of Lagrange multipliers to find the global minimum and maximum of f on the circle x2 + y2 = 4, and the points where they occur. 3. Find the global minimum and maximum of f on the disk x2 + y² < 4, and the points where they occur. Explain.

Answer 1

Solution f(wy) = 3 + ² + 3y? 2 To find out local minima local manima and food to we re points of f. first sind partial derivatives of scurg) with rexpect to and need to Smalterneaux equation x (key)=0 fy (24)=0 and y. Then we solue Hllrf, fore(y) = - 3 x 2 x + 3 u 2 = x+3x2 il fy (my) = 64 dil Now. Solving felny) = 0 fy (way) = 0 x + 3x = 0 = x (3141) = 0 il " 64=0 = y of fil Som (ii) we have x(3421) = 0 n-o X-1

tune, we get (0.0), (5.0) as critical points Now. we Define a function (ng) am Dray) = fax (9) fry (34) – [srg (14) ? here, Gual for (v.) fyg (ay) = 6 fry (rial: 0 D(y) = ((641) 6)-615 666-41) at D(R) - 6 (6441) Now. Denay) each critical points Droio) = 6(0+1) = 60 for(0,0) - 660)+1 -170 tone, Co.o) in local minima point. Now, 06-3,0) = 6(6()41) 6(+241) = -640 (-10) in a sada le points. .. and local minima value at 10.0) ix $60.o) - o A hall no local maxime points.

To use of mothod of lagrange multipliers. need to solve fx fy- xay xy2-4 gray) = x² + y 2 - 4 Now, 9-24 Cili) gy= 24 Now. from (1), Cil (illy and Girl we have 4+3 42 - 2nd -ful by = 242 - VW from Cula xxy=4 GX get and *(1+34-21) = 0 1 - 0 fi) 1+30-21=0 from (vila 64-24x = 0 24(3-1): 0 y=0 or 2-3 -A)

from (vii) and (ix), we have (0,0) from (W - 3 we get 1+34 - 9(3) = 0 1+34 = 6 3n = 5 X 5 تا ( is also a critical point. n-o (Som (vii)) Thon 4 yata Hon. (0-2) and (0,2) when Y=0 (from (ix)) Thon from Al n-t9 x 2 - 4 Hono, we get (-20) and (20) whon x = 3 (from (x)) Thom (vi) 1434-6=0 =1 - 3X35 - 5

When n = 5 یا اس Then (*) **9=4 MP4-95 9 – 36-25 9 = \ - + (-) and (ş, Honcea the points on the а (01-2), (0,2), (-2,0),(2,0), (ş,-). (şim 1 . but Cool and (5/10) does not lies on the circle for all noglect there points. now. function values at there points. global maximum function value, fwg) at contcial points global minimum points (0,-) f(0,-2) = 12 f(0, 2) f(o. 2) = 12 f(-2,0) = -2 $6-2,0) f(20) f(20) = 14 global minimum global manimum &Cş,-) +(5,-) - 19.46 (5 ST f(s. = 12.46 - 3 3 3