Math 10C - Fall 2009 - Final ExamProblem 1. [15 points]At what point (x, y, z) on the plane x + 2y − z = 5 does the minimum of the functionf(x, y, z) = x2+ 2y2+ (z + 1)2occur?Answer:Using Lagrange multipliers, we need to solve∇f = λ∇gwhereg(x, y, z) = x + 2y − z.Computing the gradients, we conclude(2x, 4y, 2(z + 1)) = λ(1, 2, −1) =⇒ x =λ2, y =λ2, z = −λ2− 1.Substituting, we havex + 2y − z = 5 =⇒λ2+ 2 ·λ2+λ2+ 1 = 5 =⇒ λ = 2.This givesx = 1, y = 1, z = −2.Problem 2. [20 points.]Consider the functionf(x, y) = 3y2− 2y3− 3x2+ 6xy.(i) [8] Find the critical points of the function.(ii) [8] Determine the nature of the critical points (local min/local max/saddle).(iii) [4] Does the function f(x, y) have a global minimum or a global maximum?Answer:(i) We havefx= −6x + 6y = 0 =⇒ x = yfy= 6y − 6y2+ 6x = 0 =⇒ y − y2+ x = 0.Substituting x = y into the second equation, we obtain2y − y2= 0 =⇒ y = 0 or y = 2.The critical points are(0, 0), (2, 2).(ii) We computeA = fxx= −6, B = fxy= 6, fyy= 6 − 12y.Whenx = y = 0 =⇒ AC − B2= (−6)(6) − 62< 0 =⇒ (0, 0) saddle point,x = y = 2 =⇒ AC − B2= (−6)(−18) − 62> 0, A < 0 =⇒ (2, 2) local max.(iii) We computelimx→∞,y=0f(x, y) = limx→∞−3x2= −∞ =⇒ no global min.Similarly,limx=0,y→−∞f(x, y) = limy→−∞3y2− 2y3= ∞ =⇒ no global max.Problem 3. [13 points] Consider the functionf(x, y) = ln(xy2) −2xy(i) [8] Compute the second order Taylor polynomial of f around (1, 1).(ii) [5] Find the tangent plane to the graph of f at the point (1, 1, −2).Answer:(i) We computef(1, 1) = ln 1 − 2 = −2fx(x, y) =y2xy2−2y=1x−2y=⇒ fx(1, 1) = −1fy(x, y) =2xyxy2+2xy2=2y+2xy2=⇒ fy(1, 1) = 4fxx= −1x2=⇒ fxx(−1, 1) = −1fxy=2y2=⇒ fxy(1, 1) = 2fyy= −2y2−4xy3=⇒ fyy(1, 1) = −6.The Taylor polynomial is−2 − (x − 1) + 4(y − 1) −12(x − 1)2+ 2(x − 1)(y − 1) − 3(y − 1)2.(ii) The tangent plane is the linear part of the Taylor polynomialz = −2 − (x − 1) + 4(y − 1) = −x + 4y − 5.Problem 4. [17 points.]Find the global minimum and global maximum of the functionf(x, y) = x2+ y2− 2x − 2y + 4over the closed diskx2+ y2≤ 8.Answer:We find the critical points in the interior by setting the partial derivatives to zero:fx= 2x − 2 = 0 =⇒ x = 1fy= 2y − 2 = 0 =⇒ y = 1.We get the critical point (1, 1) with valuef(1, 1) = 2.We check the boundary g(x, y) = x2+ y2= 8 using Lagrange mulipliers∇f = λ∇g =⇒ (2x − 2, 2y − 2) = λ(2x, 2y) =⇒ x − 1 = λx, y − 1 = λy.Dividing we obtainx − 1y − 1=λxλy=xy=⇒ y(x − 1) = x(y − 1) =⇒ xy − y = xy − x =⇒ x = y.Sincex2+ y2= 8 =⇒ x = y = ±2.We evaluatef(2, 2) = 4, f(−2, −2) = 20.Therefore (1, 1) is the global minimum, while (−2, −2) is the global maximum.Problem 5. [15 points]Consider the functionf(x, y) = 1 + x2+ y2.(i) [4] Draw the contour diagram of f labeling at least three levels of your choice.(ii) [4] Compute the gradient of f at (1, −1) and draw it on the contour diagram of part (i).(iii) [3] Does the function f have a global minimum? If no, why not? If yes, what is the minimumvalue?(iv) [4] Draw the graph of the function f.Answer:(i) The contour diagram consists of concentric circles of centered at the origin:f(x, y) = c =⇒ x2+ y2= c − 1.Level c corresponds to a circle of radius√c − 1.You may use three values for c to draw the contour diagram. For instance, for c = 2, 5, 10,we get circles of radii 1, 2, 3 respectively.(ii) The gradient is∇f = (2x, 2y) =⇒ ∇f(1, −1) = (2, −2).This vector is normal to the level curve.(iii) The global minimum occurs at (0, 0). The minimum value is f(0, 0) = 1.(iv) The graph is a paraboloid with lowest point at (0, 0, 1).Problem 6. [15 points]Consider the functionf(x, y) = e−3x+2y√2x + 1 .(i) [5] Calculate the gradient of f at (0, 0).(ii) [5] Find the directional derivative of f at (0, 0) in the direction u =i+j√2.(iii) [5] What is the unit direction for which the rate of increase of f at (0, 0) is maximal?Answer:(i) Using the product rule and the chain rule, we havefx= −3e−3x+2y√2x + 1 + e−3x+2y·12√2x + 1 · 2 =⇒ fx(0, 0) = −3 + 1 = −2,fy= 2e−3x+2y√2x + 1 =⇒ fy(0, 0) = 2.The gradient is∇f = (−2, 2).(ii) We computefu= ∇f ·u = (−2, 2) ·1√2,1√2= 0.(iii) The direction is parallel to the gradient, normalized to have unit lengthv =∇f||∇f||=(−2, 2)p(−2)2+ 22=−1√2,1√2.Problem 7. [15 points]Consider the planesx + 2y − z = 1, x + 4y − 2z = 3.(i) [4] Find normal vectors to the two planes.(ii) [6] Are the two planes parallel? Are they perpendicular?(iii) [5] Find a vector parallel to the line of intersection of the two planes.Answer:(i) The normal vectors are given by the coefficients of the two planes~n1= (1, 2, −1), ~n2= (1, 4, −2).(ii) The vectors ~n1, ~n2are not proportional hence the planes are not parallel. We have~n1·~n2= (1, 2, −1) · (1, 4, −2) = 1 + 8 − 3 6= 0.The planes are not perpendicular.(iii) The line of intersection is normal to both n1and n2hence it is parallel to the cross product~n1×~n2= (0, 1, 2).Problem 8. [15 points.]Consider the functionw = u2v e−vand assume thatu = x2− 2xy, v = −x + 2 ln y.Calculate the values of the derivatives∂w∂xand∂w∂yat the point (x, y) = (1, 1).Answer: We evaluate∂w∂x=∂w∂u·∂u∂x+∂w∂v·∂v∂x.At x = y = 1, we haveu = −1, v = −1.We compute∂w∂u= 2uve−v= 2e∂w∂v= u2e−v− u2ve−v= e + e = 2e.Furthermore,∂u∂x= 2x − 2y = 0∂v∂x= −1.Thus∂w∂x= 2e · 0 + 2e · (−1) = −2e.A similar calculation shows∂w∂y= 0.Problem 9. [10 points]You deposit $1, 000 into your savings account every year for the next 10 years. You make the first depositon January 1, 2010, and the last deposit on January 1, 2019. The interest rate for the account is r = 10%per year. How much money will there be in your account at the end of the 10thyear, on December 31, 2019?Express your answer in the simplest closed form. You don’t need to evaluate the powers that may appearin the final expression.Answer:After 1 year, on January 1, 2011 you have 1, 000 from new deposits and 1.1 ·1, 000 from depositsand interest over the previous year, a total of1, 000 + 1.1 · 1, 000.On January 1, 2012, you own1.1 · (1, 000 + 1.1 · 1, 000)from interest and 1, 000 from new deposits, a total of1, 000 + 1.1 · (1, 000 + 1.1 · 1, 000) = 1, 000 + 1.1 · 1, 000 + (1.1)2· 1,