This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MAC 2313, Fall 2010 — Midterm 3 Review Problems Solutions  1 We will discuss these problems in class on Monday 10/25. Solutions will be posted on the course webpage over the weekend. 1 Use the method of Lagrange multipliers to find the maximum and minimum values of f ( x, y ) = 3 x 4 + 5 y 4 subject to the constraint x 2 + y 2 = 1. Solution: Let g ( x, y ) = x 2 + y 2 , so the constraint is g ( x, y ) = 1. Then we have ∇ f = ( 12 x 3 , 20 y 3 ) , ∇ g = ( 2 x, 2 y ) . We want to find all triples ( x, y, λ ) which satisfy the constraint g ( x, y ) = 2 such that ∇ f = λ ∇ g , i.e., all the triples that solve the system 12 x 3 = λ 2 x, 20 y 3 = λ 2 y, x 2 + y 2 = 1 . We must be a bit careful when solving this system. If x = 0, then the first equation is automatically true for any value of λ , and to satisfy the constraint we must have y = ± 1. This gives the points (0 , 1) and (0 , − 1). Similarly, if y = 0, then the second equation is true for any value of λ and to satisfy the constraint we must have x = ± 1, giving the points (1 , 0) and ( − 1 , 0). We will evaluate f at each of these points at the end. Now suppose that neither x nor y is 0. Then we can divide both sides of the first equation by 2 x (because it is nonzero), to get λ = 6 x 2 , and we can solve this to see that x = ± radicalbig 6 / λ . From the second equation, we get y = ± radicalbig 10 / λ . Substituting these values into the constraint x 2 + y 2 = 1 gives 6 λ + 10 λ = 1 , so λ = 15 / 4 , which leads to x = ± radicalbig 5 / 8 and y = ± radicalbig 3 / 8 . We now have the following chart. Solution to Lagrange Value of f ( x , y ) multiplier equations (0 , 1) 5 max (0 , − 1) 5 max (1 , 0) 3 ( − 1 , 0) 3 ( radicalbig 5 / 8 , radicalbig 3 / 8 ) 15 / 8 min ( radicalbig 5 / 8 , − radicalbig 3 / 8 ) 15 / 8 min ( − radicalbig 5 / 8 , radicalbig 3 / 8 ) 15 / 8 min ( − radicalbig 5 / 8 , − radicalbig 3 / 8 ) 15 / 8 min MAC 2313, Fall 2010 — Midterm 3 Review Problems Solutions  2 2 Using the method of Lagrange multipliers, find the point on the surface x 2 y + zy = 2 which is closest to the origin. Solution: We want to minimize the distance to the origin, or equivalently, the square of the distance to the origin, which we call f : f ( x, y, z ) = x 2 + y 2 + z 2 . Our constraint is that the points must lie on the surface g ( x, y, z ) = 2, where g ( x, y, z ) = x 2 y + zy . We first compute the gradients: ∇ f = ( 2 x, 2 y, 2 z ) , ∇ g = ( 2 xy, x 2 + z, y ) We want to find points on the surface g ( x, y, z ) = 2 where these gradients are parallel, i.e., where ∇ f = λ ∇ g , so we need to solve the system of equations 2 x = λ 2 xy, 2 y = λ ( x 2 + z ) , 2 z = λy, x 2 y + zy = 2 Again we have to be careful. From the first equation, we see that either x = 0, or x negationslash = 0 so we can divide by it to see that y = 1 / λ . First consider the case where x = 0. The second equation becomes 2 y = λz...
View
Full Document
 Fall '08
 Keeran
 Calculus, dr dθ, Multiple integral, dx

Click to edit the document details