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UIUC MATH 241 - Worksheet_100412_sol

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Thursday, October 4 ∗ Solutions ∗ Constrained min/max via Lagrange multipliers.1. Let C be the curve in R2given by x3+y3=16.(a) Sketch the curve C.SOLUTION:(b) Is C bounded?SOLUTION:No. Given arbitrarily large y values we can find an x value which satisfies the equation. Tosee this notice that y =3p16 −x3, so we can input arbitrarily large (or small) x values andget a y value for that input.(c) Is C closed?SOLUTION:Yes, C is closed in R2.2. Consider the function f (x, y) =ex yon C.(a) Is f continuous? What does the Extreme Value Theorem tell you about the existance ofglobal min and max of f on C?SOLUTION:Yes, f is continuous. Since C is not bounded, the Extreme Value Theorem does not tell youanything about the existence of a global min and max of f on C.(b) Use Lagrange multipliers to determine both the min and max values of f on C.SOLUTION:Let g (x, y) = x3+y3. Our constraint is g (x, y) =16. ∇f =(yex y, xex y) and ∇g = (3x2,3y2),so using the method of Lagrange multipliers we need to find simultaneous solutions in xand y of the following three equations:x3+y3= 16 (1)yex y= λ3x2(2)xex y= λ3y2(3)Multiplying (2) by x gives x yex y= λ3x3and multiplying (3) by y gives yxex y= λ3y3. Sowe have that λx3= λy3. This is satisfied if λ = 0 or if x3= y3. If λ = 0 we deduce from (2)that y = 0 and from (3) that x = 0. But the point (0,0) is not on the curve x3+y3= 16, soλ 6= 0. So we must have x3= y3, or x = y. Using (1) this implies that 2x3= 16 or x = y = 2.So f attains either a maximum or a minimum of f (2,2) =e4at (2,2).I claim f (2,2) =e4is the global maximum of f on C . One way to see this is that since f hasonly one critical point on C , it must behave in one of exactly two ways:i. f increases on C as x increases until it hits x =2, then f decreases. In this case f hasa global maximum at (2,2).ii. f decreases on C as x increases until it hits x = 2, then f increases. In this case f hasa global minimum at (2,2).From the graph of x3+ y3= 16 we see that most of C lies in either the second or fourthquadrant, implying that x y <0 on most of C , or ex y<1. Since e4>1, we see that f cannothave a global minimum at (2, 2), so it must have a global maximum there. Since there is noother critical point, f does not have a minimum on C . In fact we can make f arbitrarilyclose to 0 by taking points on C with either very large or very small x coordinate.3. Consider the surface S given by z2= x2+y2(a) Sketch S.SOLUTION:(b) Use Lagrange multipliers to find the points on S that are closest to (4,2,0).SOLUTION:Minimize the square of the distance function D = (x −4)2+(y −2)2+z2from the point(4,2, 0) subject to the constraint g = x2+y2−z2= 0. We have ∇D =2(x −4),2(y −2),2z®and ∇ g =2x,2y, −2z®. From the picture it is clear that D attains a global minimum valueon S (i.e. there are points which are closest to (4,2,0)). So one of the critical points we findusing Lagrange multipliers will correspond to this minimum value and we simply needto evaluate D at each of the critical points and take the smallest to find the minimumdistance. Using the method of Lagrange multipliers we get the system (divide out by 2first):(x −4) =λx(y −2) =λyz =−λ zIf λ 6=−1 then z =0 from the last equation so the constraining equation z2= x2+y2impliesthat x = y = 0. If λ = −1 then the top two equations give x = 2 and y = 1. So the threepossible points of minimum distance from (4,2,0) are (0,0, 0), (2,1,p5), and (2, 1,−p5). Bycalculation we see that the squares of the distances of each of these from (4,2, 0) are 20,10,and 10 respectively. So the two points (2,1,p5) and (2,1,−p5) on the cone z2= x2+y2areof minimum distance from the point (4,2,0).4. For the function shown on the back of the sheet, use the level curves to find the locations andtypes (min/max/saddle) for all the critical points of the function:f (x, y) =3x −x3−2y2+y4Use the formula for f and the 2nd-derivative test to check your answer.SOLUTION:Mins and maxes occur where the level curves shrink toward a point and saddle points occurwhere the level curve intersects itself. From looking at the set of level curves it appears thatf (x, y) has minimums at (−1,1) and (−1, −1), a maximum at (1,0), and saddle points at (−1,0),(1,1), and (1, −1).Now let’s find the critical points precisely. fx= 3(1 −x2) and fy= 4y(y2−1). So f has criticalpoints at (1,0),(1, 1),(1,−1),(−1,0),(−1,1), and (−1,−1). fxx= −6x, fy y= 12y2−4, and fx y= 0,so the Hessian is D(x, y) = fxxfy y−( fx y)2=−6x(12y2−4). D(−1,0),D(1,1), and D(1, −1) are allnegative, so these are saddle points. D(1, 0),D(−1, 1), and D(−1,−1) are all positive so these aremaxes and mins. fxx(1,0) <0 so (1,0) is a local max. fxx(−1,1) and fxx(−1,−1) are both positiveso these are local mins. This analysis agrees with our guesses.5. If the length of the diagonal of a rectangular box must be L, what is the largest possible volume?SOLUTION:Set x =length of the box, y = width of the box,z =height of the box. This simply supposes thatthe box is sitting in the octant x ≥ 0, y ≥0, and z ≥0 with its edges along each axis. The volumefunction is then V = x yz and the constraint is that L2= x2+ y2+ z2. Using the method ofLagrange multipliers we get the system of equations:yz =2λxxz =2λyx y =2λzx2+y2+z2=L2Since we want to maximize volume we can assume that x > 0, y > 0, and z > 0. This rules outthe possibility λ = 0 (since λ =0 implies at least two of the variables x, y, and z are 0). Also thismeans we can multiply the first equation by x, the second by y, and the third by z to get a newsystem:x y z =2λx2x y z =2λy2x y z =2λz2This implies that x2= y2= z2. Coupling this with the constraints x > 0, y > 0, z > 0 we see thatthis means x = y = z. Plugging this into the constraining equation L2= x2+y2+z2we get thatL2=3x2or x =L/p3. So V =(L/p3)3=L3/(3p3) is the biggest possible volume for the


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UIUC MATH 241 - Worksheet_100412_sol

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