Question

Problem 5. Find the local marimum and minimum values and saddle point(s) of the functions: i) f(x,y) = x2 + xy + y2 + y. a) f(x, y) = (x - y)(1 - x). ui) (Optional) f(0,y) = xy +e-zy. Note that the critical points are (2,0) and (0,y) and that f(x,0) = f(0, y) = 1. However, from Math 110, we can show that the function gw) = w+e-w has an absolute mim at w = 0i.e., g(w) > g(0) = 1 for all w. Therefore, f(x,y) = xy + e- > 1 for all (2,y). Thus, f has a local min at any point (2,y) on 2- or y-acis. Problem 6. Introduction to the problem: Recall that the distance be- tween two points (21,41, 21) an (22,42, 22) in R' is given by d= (x2 - 31)2 + (y2 - yı)2 + (22 – 21) For example, the distance between a point (co, yo, 20) in R3 and any point (:, y, z) on a plane ar+ay+cz = k is equal to the function of two variables: d = f(x,y) = (x – xo)2 + (1 – yo)2 + ((_..y) – zo)" Note that, since (2, y, z) lies on the plane then it has to satisfy its equation. i.e., z = - Note also that 0 <d always holds, and suppose that f(cy) has a local min at the point (a,b). Then, by definition 0 <f(a,b) <f(x,y) for all (x, y) near (a,). Therefore, if we square both sides, we get 0 <f(a, B) <f(a, y) for all (2,y) near (a, b). That means, (a, b) is also a local minimum for the function . The problem itself: Find the shortest distance from the point (1,0,-2) to the plane x + 2y + z = 4. Problem 7. Evaluate the double integral S(:2 - 3y2) dA, where R= {(x,y) 0<x<2, 1<y <2}

Answer 1

Ow The given function is - f(x,y) = x + aytyrty First we find the partial derivatives - fx = 28+7 fy = 24+ 2+1 fxx=2, tyg 2 . fag=1, two=1 We know that if (a, b) be the stationary point, then fa 19,6)=0=fy (9,6). and if fun fyy- tayfyr so at (a,b), then (a, b) is à saddle point. *If free tyy- tantre yo at (a,b), then Ab) is either minimum or maximum point. Distinguish between these as follows- *if tax so and tyg so at (2,6), then (a,b) is a maximum point. * If fra 20 and fun to at (9,6), then (9,6) is a minimum pont.

Now for the given problem, the stationary points are obtained as follows - toto and fy co gives 2x+y=0 24+x+1=0 y=-2* 08, 2 (-22) +x+10 (as 2=-2x) O, -3x+120 or, x iz This (3-2) is a stationary point of f(x,y). New at (6-3) - fxx Syge fuiy ty=4-1=350 Thus E-2) is an extremum point sile, either minimum or manimum. But there saddle point. does not exist any Now we check whether the point (6-3) io marimum og minimum. Now at (3,3);- Fox=270 fyy=270. Thus (0-3) is a minimum pont. Hence the local menimum value of f(x,y) in 1(3-3)= ($)+ (%)(-3) + (33)-(3) - - Św (ii) there fox,y) = (2-13) (1-xy) ie 1083) = x-xy-y+ey Therefore sa 1-2xy + yt - - + 2x - c1 - 27,2x

fay= 27- 2x fyx= 23-2x The stationary points are obtained by fo=0 and fy=0 ie 1-2xy + y = 0 2xy-x-1=0 or, 1-27: -yr or, 1-242 +120 os eny=0 (as 1-2xy=-yn 2: I 1- 2x2 + 3%20 or x + 2x +130 (prit x2.7) or, 2= is invalid Here any since, fe to gives 1-2xy +Yr=0 -) 2+2x+120 , (as x2-y) je 43x43- er, a -ţ , (an impossible cove) Thus ry. and the statimary point is (1,1),(-1,-1) Now at (11) den Fry- (fey) = (-2). (2) - (2-2) = – 4-02-450 Hence (1.1) is a saddle point.

Now at (1,1). fox fyy-Grey) = (2) (-2) - (-2+2) --4-0--440 Hence (-1,-1) is also a saddle point. Therefore minimum and manimum valce does not exist for this given function Shortest distance from the point (1,0,-2) to the plane x+24+2=4. We know that if the equation of a plane be ax+byt (2+0 =0, then the perpendicular distance i e, Shortest distance of the point (16,7,2) from the plane (above). is given by at axi+by,+ CZ, Jateter: We are to take (tive. Sign if dyo, we are to take the sign if d <o: Therefore for the given plane, the required Shortest dis same is 1 (1) + 2(0)+1(-2) 1+ 4+1 :-( to unit te mit cenit, (as dind here)