First point ( 0 , 0 ) of type S Second point ( 0 , __ ) of type S Third point (
ID: 2832455 • Letter: F
Question
First point ( 0 , 0 ) of type S
Second point ( 0 , __ ) of type S
Third point ( __ , __ ) of type MA
Fourth point ( __ , 0 ) of type S
(i filled out what i already solved)
2)
-------------------------------------------------------------
3)
4)
5)
Use Lagrange multipliers to find the coordinates of the point (x, y, z) on the plane z = 1 x + 3 y + 4 which is closest to the origin.
x =
y =
z =
has 4 critical points. List them in increasing lexographic order. By that we mean that (x, y) comes before (z, w) if or if and . Also, describe the type of critical point by typing MA if it is a local maximum, MI if it is a local minimim, and S if it is a saddle point.
First point ( 0 , 0 ) of type S
Second point ( 0 , __ ) of type S
Third point ( __ , __ ) of type MA
Fourth point ( __ , 0 ) of type S
(i filled out what i already solved)
Explanation / Answer
Answer 1)
First point ( 0 , 0 ) of type S
Second point ( 0 , 0.25 ) of type S
Third point ( 0.1111 , 0.08333 ) of type MA
Fourth point (0.3333 , 0 ) of type S
Answer 2)
Given: Volume, V = xyz = 4913
Surface area, S = 2xy + 2xz + 2yz
Take partial derivatives. To find the minimum, they would each be set to zero,
but that means they will be equal to each other.
Partial derivative of S wrt x is: ?S/?x = 2(y + z)
Partial derivative of S wrt y is: ?S/?y = 2(x + z)
Partial derivative of S wrt z is: ?S/?z = 2(x + y)
Therefore, 2(y + z) = 2(x + z) = 2(x + y)
or, y + z = x + z = x + y
Taking any 2 equations, three times, we find that y = x, z = x and z = y.
In other words, x = y = z is the solution for a minimum, so the box is a cube.
Substituting y = x and z = x gives: V = x^3 =8000, so, x = 8000^(1/3) = 20
i.e. x = 20, y = 20, z = 20, yielding S = 3*(2*20*20) = 2400.
Answer 3)
Set the equation equal to zero.
z = 3x + 2y + 4
3x + 2y - z + 4 = 0
v = n = <3, 2, -1>
The equation of the line is:
r(t) = O + tv
r(t) = <0, 0, 0> + t<3, 2, -1>
The parametric equations of the line are:
r(t):
x = 3t
y = 2t
z = -t
Substitute the values for the variables into the equation of the plane and solve for t.
3x + 2y - z + 4 = 0
3(3t) + 2(2t) - (-t) + 4 = 0
9t + 4t + t = -4
14t = -4
t = -2/7
Plug the value for t into the parametric equations of the line and solve for the closest point in the plane to the origin.
r(t):
x = 3(-2/7) = -6/7
y = 2(-2/7) = -4/7
z = -(-2/7) = 2/7
The closest point on the plane to the origin is P(-6/7, -4/7, 2/7).
Answer 4)
Minimize x^2+y^2+z^2, with restraint z=x+3y+4.
Introduce L (for Lambda), a 4th unknown. Then at the minimum Grad(x^2+y^2+z^2) = L*Grad(x+3y+4-z) and z=x+3y+4.
Taking the gradients leads to 3 equations (1 per coordinate) and the last =x+3y+4 gives 4 equations for the 4 unknowns.
(2x, 2y, 2z) = L(1, 3, -1), and z=x+3y+4.
So x=L/2, y=3L/2, z=-L/2, and z=x+3y+4.
Substituting Ls gives
-L/2=L/2+3*3L/2+4
L=-8/11
(x,y,z) = (-4/11, -12/11, 4/11)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.