Math 233. Lagrange Multipliers Basics

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1 Math 33. Lagrange Multipliers Basics Optimization problems of the form to optimize a function f(x, y, z) over a constraint g(x, y, z) = k can often be conveniently solved using the method of Lagrange multipliers: Theorem. Let f and g have continuous first partial derivatives and suppose f has an extremum at (x 0, y 0, z 0 ) on the smooth surface g(x, y, z) = c. If g(x 0, y 0, z 0 ) is not the zero vector, then there is a real number λ such that f(x 0, y 0, z 0 ) = λ g(x 0, y 0, z 0 ) Analogs of this hold whenever f and g are functions of the same number of variables ( or more). Geometrically this means the gradients of f and g are parallel when f attains an extreme value on a level surface of g. Under the conditions of the theorem, the relation f = λ g will be true when f has a maximum or a minimum on the surface, however, the condition f = λ g at a point does not guarantee f has a maximum or minimum on at a point on the level surface of g, the only way to determine maximum and minimum values of f on the surface would be to test all possible points where the gradients are parallel while making sure to that maximums and/or minimums actually exist. Example 1. A certain company s shipping regulation is that the sum of the length and girth (perimeter of a cross-section) of a rectangular package cannot exceed 138 inches. Find the dimensions of the rectangular package of largest volume that may be shipped. Answer. For a box with sides x, y and length z the girth is x + y and the length is z. We then maximize volume f(x, y, z) = xyz subject to the constraint g(x, y, z) = x+y +z = 138. The equation f = λ g leads to the equations yz = λ xz = λ xy = λ This implies λ = yz = xz and so y = x (since for maximal volume x 0, y 0 and z 0) and λ = yz = xy and so z = x. Plugging these relations in the constraint x + y + z = 138 result in x + x + x = 138 = 6x = 138 = x = 3 Then z = (3) = 46 and the dimensions of the package are 3 inches by 3 inches by 46 inches yielding a maximum volume of 4334 cubic inches. Example. A cargo container (in the shape of a rectangular solid) must have a volume of 600 cubic feet. The bottom needs to be stronger, so it will cost $6 per square foot to construct, whereas the sides and top will cost $3 per square foot to construct. Find the dimensions of the container of this size that has minimum cost. Answer. Minimize cost: f(x, y, z) = 9xy + 6xz + 6yz (x, y are lengths of edges of base, z is height)

2 Subject to g(x, y, z) = xyz = 600. Using Lagrange multipliers we get 9y + 6z = λyz, 9x + 6z = λxz, 6x + 6y = λxy and because of the constraint, x 0, y 0 and z 0. Thus solving each of these for λ and equating them pairwise we get 9 z + 6 y = 9 z + 6 x 9 z + 6 x = 6 y + 6 x. The first equation above implies x = y and the second implies z = 3 y. Plugging these into the constraint we have y y 3 y = 600 implies y3 = 400. Thus the dimensions of the box are x = 3 400, y = 3 400, and z = Example 3. Find the point on the plane 3x y + z = 8 that is closest to (0, 0, 0). Answer. Minimize f(x, y, z) = x + y + z subject to g(x, y, z) = 3x y + z = 8. Using Lagrange multipliers one solves f = λ g, or x = 3λ y = λ z = λ The first and third equation imply x = 3z; the second and third imply y = z. When these relations are substituted back into the constraint we obtain 9z + 4z + z = 8. This implies z =. The nearest point is thus (6, 4, ). Example 4. (A Lagrange multiplier question with two constraints.) Find the maximum and minimum volumes of a rectangular box along with their corresponding dimensions whose surface area equals 9000 square cm and whose edge length (sum of lengths of all twelve edges) is 50 cm. Answer. First, the edges cannot all be equal, because if they were, the edge length would be 50/1 = 43 cm and then the surface area would be 6(43) = square cm which exceeds the given value of 9000 square cm for surface area. Thus, there are at most two sides with equal length, we will label these sides y and z, with the third side being x. Thus we know, x y and x z. Now we set up the Lagrange multiplier equations: We will maximize and minimize V = xyz subject to the constraints xy + xz + yz = 9000 and 4x + 4y + 4z = 50 We divide these constraint equations by and 4 respectively to obtain the Lagrange multiplier problem Maximize and minimize f(x, y, z) = xyz subject to g(x, y, z) = xy + xz + yz = 4500 and h(x, y, z) = x + y + z = 130

3 Then f = λ g + µ h implies yz = λ(y + z) + µ xz = λ(x + z) + µ xy = λ(x + y) + µ Subtracting the second equation from the first above yields yz xz = λy λx = z(y x) = λ(y x) = z = λ since y x Similarly, subtracting the third equation from the first above implies y = λ. Thus we conclude y = z. Then the constraint x + y + z = 130 = x + y = 130 = x = 130 y Finally, putting this information into the constraint xy + xz + yz = 4500 yields (130 y)y + (130 y)y + y = 4500 = 3y 60y = 0 The quadratic formula then implies y = 60 ± 60 4(3)(4500) 6 = y or y In the minimum volume case, the volume is approximately cm 3 and the dimensions are approximately cm by cm by cm In the maximum volume case, the volume is approximately cm 3 and the dimensions are approximately cm by cm by cm Example 5. The surface are of a right-circular cone of radius r and height h is S = πr r + h, and its volume is V = 1 3 πr h. Find the radius and height of a cone of maximum volume whose surface area is S = 8. Answer. We will use Lagrange multipliers to maximize f(r, h) = 1 3 πr h subject to g(r, h) = πr r + h = 8 Let f r = λg r and f h = λg h yields 3 πrh = λ ( π r + h + ) r r + h and 1 πrh 3 πr = λ r + h we know r 0 at a maximum volume, and so the previous two equations simplify to ( ) r 3 rh = λ r + h + and 1 r + h 3 r = λ πrh r + h Solving the second equation for λ we have λ = r r + h 3h

4 Now substitute this expression for λ in to the first equation to obtain 3 rh = r ( ) r + h r r + h 3h + or r + h 3 rh = r 3h [(r + h ) + r ] and so multiplying both sides by 3h and dividing by r implies h = (r + h ) + r which implies h = r and so h = r Putting h = r in the constraint implies 8 = πr 3r and so r = 8/(π 3) 1.15 and h = (1.15)

5 Practice Exercises on Extrema 1. A cargo container (in the shape of a rectangular box) must have a volume of 4608 cubic feet. The bottom needs to be stronger, so it will cost $5 per square foot to construct, whereas the sides and top will cost $4 per square foot to construct. Find the dimensions of the container of this volume that has minimum cost.. Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid x 81 + y 36 + z 9 = 1 3. Find the point on the plane 4x 5y + 5z = 13 that is closest to (0, 0, 0). 4. (A Lagrange multiplier question with two constraints.) Find the maximum and minimum volumes of a rectangular box along with their corresponding dimensions whose surface area equals square cm and whose edge length (sum of lengths of all twelve edges) is 564 cm. 5. A certain company s shipping regulation is that the sum of the length and girth (perimeter of a cross-section) of a rectangular package cannot exceed 108 inches. Find the dimensions of the rectangular package of largest volume that may be shipped. 6. Find the minimum and maximum values of the function on the domain x + y 90. f(x, y) = 3x + 4y 6x 1 7. Find the maximum and minimum values of f(x, y) = 13x+y on the ellipse 6x +1y = 1. Express answers to four decimal places. 8. Find the radius and height of a closed tin can of maximal volume that is in the shape of a right circular cylinder and whose surface area is 16π square centimeters (the surface area includes the side wall, and top and bottom).

6 Practice Exercises on Extrema with Solutions. 1. A cargo container (in the shape of a rectangular box) must have a volume of 4608 cubic feet. The bottom needs to be stronger, so it will cost $5 per square foot to construct, whereas the sides and top will cost $4 per square foot to construct. Find the dimensions of the container of this volume that has minimum cost. Solution: Let x, y be lengths of edges of base, and let z be the height measured in feet. Then we Minimize cost: f(x, y, z) = 9xy + 8xz + 8yz Subject to g(x, y, z) = xyz = Using Lagrange multipliers we get 9y + 8z = λyz, 9x + 8z = λxz, 8x + 8y = λxy and because of the constraint, x 0, y 0 and z 0. Thus solving each of these for λ and equating them pairwise we get 9y + 8z yz = 9x + 8z xz = 9xy + 8xz = 9xy + 8yz = x = y Similarly, 9y + 8z yz = 8y + 8x xy = 9xy + 8xz = 8yz + 8xz = z = 9 8 x Plugging these into the constraint we have x x 9 8 x = 4608 = x3 = 4096 = x = 16 Thus the dimensions of the box are x = 16 feet, y = 16 feet, and z = 18 feet. That is, the base is 16 feet by 16 feet, and the height is 18 feet.. Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid x 81 + y 36 + z 9 = 1 Solution: We will maximize f(x, y, z) = xyz subject to the constraint g(x, y, z) = x 81 + y 36 + z 9 = 1 where x, y, z are in the first quadrant, then the maximum volume will be 8 times the maximum value for f. Then setting f = λ g we obtain yz = λx 81 xz = λy 36 xy = λz 9

7 Notice that for a maximum, x 0, y 0 and z 0. Then dividing the first equation by the second we have yz xz = λx λy = y x = 36x 81y y = x = = y = 36x 81 Similary, dividing the first equation by the third we have yz xy = λx 81 9 λz = z x = 9x 81z z = x = 9 81 = z = 9x 81 Substituting these into the constraint, we find then y = 6 9 x x (81)(36) + 9x (81)(9) = 1 = x = 81 3 = x = = 6 and z = = 3. Then the maximum volume is 3 3 V = = Find the point on the plane 4x 5y + 5z = 13 that is closest to (0, 0, 0). Solution: Minimize f(x, y, z) = x +y +z subject to g(x, y, z) = 4x 5y+5z = 13. Using Lagrange multipliers one solves f = λ g, or Then and so y = 5x, and z = 5 x and then 4 4 x = 4λ y = 5λ z = 5λ λ = x 4 = y 5 = z 5 4x x + 5 x = 13 = 66x = 58 = x = 8 4 Then y = 5 4 (8) = 10 and z = 5 (8) = 10. Thus the point on the plane nearest to 4 (0, 0, 0) is (8, 10, 10). 4. (A Lagrange multiplier question with two constraints.) Find the maximum and minimum volumes of a rectangular box along with their corresponding dimensions whose surface area equals square cm and whose edge length (sum of lengths of all twelve edges) is 564 cm.

8 Solution: First, the edges cannot all be equal, because if they were, the edge length would be 564/1 = 47 cm and then the surface area would be 6(47) = 1354 square cm which exceeds the given value of square cm for surface area. Thus, there are at most two sides with equal length, we will label these sides y and z, with the third side being x. Thus we know, x y and x z. Now we set up the Lagrange multiplier equations: We will maximize and minimize V = xyz subject to the constraints xy + xz + yz = and 4x + 4y + 4z = 564 We divide these constraint equations by and 4 respectively to obtain the Lagrange multiplier problem Maximize and minimize f(x, y, z) = xyz subject to g(x, y, z) = xy + xz + yz = 6550 and h(x, y, z) = x + y + z = 141 Then f = λ g + µ h implies yz = λ(y + z) + µ xz = λ(x + z) + µ xy = λ(x + y) + µ Subtracting the second equation from the first above yields yz xz = λy λx = z(y x) = λ(y x) = z = λ since y x Similarly, subtracting the third equation from the first above implies y = λ. Thus we conclude y = z. Then the constraint x + y + z = 141 = x + y = 141 = x = 141 y Finally, putting this information into the constraint xy + xz + yz = 6550 yields (141 y)y + (141 y)y + y = 6550 = 3y 8y = 0 The quadratic formula then implies y = 8 ± 8 4(3)(6550) 6 = y or y In the minimum volume case, the volume is approximately cm 3 and the dimensions are approximately cm by cm by cm In the maximum volume case, the volume is approximately cm 3 and the dimensions are approximately cm by cm by cm

9 5. A certain company s shipping regulation is that the sum of the length and girth (perimeter of a cross-section) of a rectangular package cannot exceed 108 inches. Find the dimensions of the rectangular package of largest volume that may be shipped. Solution: For a box with sides x, y and length z the girth is x + y and the length is z. We then maximize volume f(x, y, z) = xyz subject to the constraint g(x, y, z) = x + y + z = 108. The equation f = λ g leads to the equations yz = λ xz = λ xy = λ This implies λ = yz = xz and so y = x (since for maximal volume x 0, y 0 and z 0) and λ = yz = xy and so z = x. Plugging y = x and z = x in the constraint x + y + z = 108 results in x + x + x = 108 = 6x = 108 = x = 18 Then z = (18) = 36 and the dimensions of the package are 18 inches by 18 inches by 36 inches yielding a maximum volume of cubic inches. 6. Find the minimum and maximum values of the function on the domain x + y 90. f(x, y) = 3x + 4y 6x 1 Solution: We will consider the problem to find the maximum and minimum of f(x, y) subject to the constraint g(x, y) = x + y = c where c is any number from 0 to 90. Then f(x, y) = λ g implies 6x 6 = λx and 8y = λy In the case y 0, the equation 8y = λy implies λ = 4. Plugging λ = 4 back into the first equation implies 6x 6 = 8x and so x = 3. This leads to a maximum when c = 90 and there is no minimum since c 0. In this case x = 9 and y = 81 and so f(x, y) = 375 which is the maximum value of f achieved on the given domain. In the case y = 0, we have f(x, 0) = 3x 6x 1. This has a minimum when 6x 6 = 0 and so x = 1. This minimum is f(1, 0) = 3(1 ) (6)(1) 1 = 15 which is the minimum value achieved by f on the given domain. 7. Find the maximum and minimum values of f(x, y) = 13x+y on the ellipse 6x +1y = 1. Express answers to four decimal places.

10 Solution: We maximize f(x, y) subject to the constraint g(x, y) = 6x + 1y = 1. The equation f = λ g implies 13 = (6)λx and = (6)λy These equations imply λ 0, and so we solve for x and y obtaining x = 13 (6)λ and y = (1)λ Plugging these values back into the constraint we have 13 4(6)λ + 4(1)λ = 1 = λ = ± 13 4(6) + 4(1) That is λ ± and so substituting back into equations for x and y we find The maximum is approximately x ± and y ± f( , ) = (13)( ) + ()( ) and the minimum is approximately f( , ) = (13)( ) ()( ) Find the radius and height of a closed tin can of maximal volume that is in the shape of a right circular cylinder and whose surface area is 16π square centimeters (the surface area includes the side wall, and top and bottom). Solution: Let r denote the radius and let h denote the height of the can. Then we maximize V (r, h) = πr h subject to S(r, h) = πr + πrh = 16π Using Lagrange multipliers we solve V = λ S, thus The second equation implies πrh = λ(4πr + πh) and πr = λ(πr) r = λ(r) = λ = r since r 0 in the maximum volume case. Plugging this in the first equation implies πrh = πr + πrh = rh = r = h = r since r 0.

11 We put h = r into the constraint to find πr + πr(r) = 16π = 6πr = 16π = r = 36 = r = 6 Thus the can has radius r = 6 cm and height h = 1 cm producing a maximum volume of V = 43π cubic cm.

Math 233. Lagrange Multipliers Basics

Math 233. Lagrange Multipliers Basics Optimization problems of the form to optimize a function f(x, y, z) over a constraint g(x, y, z) = k can often be conveniently solved using the method of Lagrange