Question

Using the method of Lagrange Multipliers, find the point on the plane x+y−z=1 that is closest to the point (0, −2, 1).

Answer #1

Use Lagrange multipliers to find the point on the given plane
that is closest to the following point. (Enter your answer as a
fraction.) x - y + z = 2; (7, 7, 1)

Use Lagrange multipliers to find the point on the
plane
x − 2y + 3z = 6
that is closest to the point
(0, 2, 5).
(x, y, z) =

1. Use the method of Lagrange multipliers to find the
maximize
of the function f (x, y) = 25-x^2-y^2 subject to the constraint
x + y =-1
2. Use the method of Lagrange multipliers to find the
minimum
of the function f (x, y) = y^2+6x subject to the constraint
y-2x= 0

Using Lagrange multipliers, find the coordinates of the minimum
point on the graph of z=x2+y2 subject to the constraint
2x+y=20.
Lagrange function (use k for lambda) L(x,y,k)=
Lx(x,y,k)=
Ly(x,y,k)=
Lk(x,y,k)=
Minimum Point (format (x,y,z)):

(Lagrange Multipliers with Three Variables) Find the global
minimum value of f(x,y,z)=(x^2/4)+y^2 +(z^2/9) subject to x - y + z
= 8. Now sketch level surfaces f(x,y,z) = k for k = 0; 1; 4 and the
plane x-y +z = 8 on the same set of axes to help you explain why
the point you found corresponds to a minimum value and why there
will be no maximum value.

Use the method of Lagrange Multipliers to find the maximum
value:
f(x,y,z) = x2y2z2 subject to
the constraint x2+y2+z2=1 no
decimals permitted

use Lagrange multipliers to find the maximum of F(x,y,z)=xyz
with the constraint H(x,y,z)=x^2+y^2+z^2-8=0

Use Lagrange multipliers to find the highest point on the curve
of intersection of the surfaces. (double-check your
answer!)
Sphere: x2 + y2 + z2 =
30, Plane: 2x + y − z = 4
(x, y, z) =

Use the method of Lagrange multipliers to find the minimum value
of the function
f(x,y,z)=x2+y2+z2
subject to the constraints x+y=10 and 2y−z=3.

Use Lagrange multipliers to find the highest point on the curve
of intersection of the surfaces.
Sphere: x2 + y2 + z2 =
24, Plane: 2x + y − z = 2

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