Respuesta :
Answer:
a)
Highest (-3,-3)
Lowest (2,2)
b)
Farthest (-3,-3)
Closest (2,2)
Step-by-step explanation:
To solve this problem we will be using Lagrange multipliers.
a)
Let us find out first the restriction, which is the projection of the intersection on the XY-plane.
From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:
[tex]\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12[/tex]
completing the squares:
[tex]\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2[/tex]
and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of
[tex]\bf f(x,y)=x^2+y^2[/tex]
subject to the constraint
[tex]\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0[/tex]
Now we have
[tex]\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)[/tex]
Let [tex]\bf \lambda[/tex] be the Lagrange multiplier.
The maximum and minimum must occur at points where
[tex]\bf \nabla f=\lambda\nabla g[/tex]
that is,
[tex]\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)[/tex]
we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so
[tex]\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y[/tex]
Replacing in the constraint
[tex]\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2[/tex]
from this we get
x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3
and the candidates for maximum and minimum are (2,2) and (-3,-3).
Replacing these values in f, we see that
f(-3,-3) = 9+9 = 18 is the maximum and
f(2,2) = 4+4 = 8 is the minimum
b)
Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.
We have then,
(-3,-3) is the farthest from the origin
(2,2) is the closest to the origin.
The minimum distance of the plane will be 3✓11/2 and the maximum distance will be 6✓2.
How to calculate the distance?
The distance between a point (x, y, z) and the origin is ✓x² + ✓y² + ✓z².
In this case, we are looking for the extrema subject to x + y + 2z = 12 and z = x² + y².
The lagrangian will be:
L(x, y, z, λ, u) = x² + y² + z² + λ(x + y + 2z - 12) + u(z - x² - y²)
The partial derivatives will be:
Lx = 2x + λ - 2ux = 0
Ly = 2y + λ - 2uy = 0
Lz = 2z + 2λ + u = 0
Lλ = x + y + 2z - 12 = 0
Lu = z - x² - y² = 0
From the first two equations, it follows rhat x = y. In the last two equations, x + y + 2z - 12 = 0
Therefore, x + z = 6
x + 2x² = 6
2x² + x - 6 = 0
= (2x - 3)(x + 2)
x = 3/2 or -2.
The critical points will be (3/2, 3/2,9/2) and (-2, -2, 8)
Therefore, the minimum distance will be 3✓11/2 and the maximum distance will be 6✓2.
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