What is the maximum value of f(x,y,z) on the given conditions?

[ori]

question:
max value of f=x+y+z
on the ball x^2+y^2+z^2=a^2 (not inside the ball, only on it)
a>0
with contidions
x<=0
y<=0
z<=0
is equal to?

i get -sqrt(3)a
the right answer is -a

help me please

i'll say what I've done:
grad(x+y+z)+j*grad(x^2+y^2+z^2-a^2)=(0,0,0)
x^2+y^2+z^2=a^2
(j symbolies gama)

for example from the x equation we get:
1+2jx=0

x=-1/(2j)

same with y&z

when we put those x,y,z at the ball equation we get
j^2=3/(4a^2)

j=sqrt(3)/(2a)
therefore:
x=y=z=-1/(2j)=-a/sqrt(3)
therefore
max value: -sqrt(3)*a
where's my mistake?

 

[arildno]

Note that they're asking for max value, not min value.
Hence, you must check on the boundary of your region.
(f=-a is achieved on either x,y,z=-a, the other two zero)

 

[HallsofIvy]

The Lagrange multiplier method clearly gives x= y= z and, since we require that (x,y,z) be on the surface of the sphere with radius a as well as requiring that x,y,z be non-negative , 3x²= a² so $x= y= z= -\frac{a}{sqrt{3}}, just as you say. And, as arildno said, that is a minimum not a maximum. You will now have to look for a maximum on the boundary of that set: the three curves<br /> x^{2^{+y²= a², z= 0, z^{2^{+y²= a², x= 0, and x^{2^{+z²= a², x= 0 as well at the three points where those curves intersect, (1,0,0), (0,1,0), and (0,0,1).}}}}}}$

 

[ori]

i got it
thank u both
but i have another question with Lagrange multiplier method :
i need to find on this closed curve: x^(2/5)+y^(2/5)=1
the points that the func f(x,y)=64x-27y gets its extrime values
there's some subsections at the question, and i didnt understand them all:
the first question is what is the number of points the are candidate to be solutions of this extreme problem
the 2nd question is what is the num of pts that solve lagrange multiplier equations of this extreme problem
i don't understand what the diffrence between the questions - unlike the 1st question of this thread here we don't have stopping contiodion on the curve (like at the other question we had z<=0 etc)
another thing is that the solution of lagrange multiplier equations we get is with ugly number, and though i can understand why there's two points that solve the lagrange equations, its very ugly to get what are they
and i afraid i need them for next sub-questions
another subquestion is - at a point that solve lagrange equations of this prob , the level line of f(x,y) (which means , i think f(x,y)=const ) and the condition curve:
x^(2/5)+y^(2/5)=1 are:
1. cut with 90deg between them
2. don't cut
3. tangent each other
4. its impossible to know deg of the cutting coz the condition curve don't have tangent at this point
please help
thanks to any helper

 

[HallsofIvy]

I'm not sure just what question 1 intends: I would be inclined to say that "candidate points" are all points on the figure.
For any number F, 64x-27y = F is a straight line with slope 64/27 and increasing F moves it moves it down. The graph of x^(2/5)+y^(2/5)=1 is a "hyper-ellipse" (if I remember the name correctly) and looks like a diamond with sides sagging inward. Since moving the straight line up or down decreases and increases F, the extreme values will be where moving slightly one way or the other would take the OUT of the "box" formed by the closed curve. IF that box were smooth(i.e. did not have "corners"), that would be where then line is tangent to the curve. However, here it looks like the extreme values will be at the "corners", (1,0), (0,1), (-1,0), and (0,1) where the curve does not have a tangent: answer 4.

 

[ori]

I'm not sure just what question 1 intends: I would be inclined to say that "candidate points" are all points on the figure.
For any number F, 64x-27y = F is a straight line with slope 64/27 and increasing F moves it moves it down. The graph of x^(2/5)+y^(2/5)=1 is a "hyper-ellipse" (if I remember the name correctly) and looks like a diamond with sides sagging inward. Since moving the straight line up or down decreases and increases F, the extreme values will be where moving slightly one way or the other would take the OUT of the "box" formed by the closed curve. IF that box were smooth(i.e. did not have "corners"), that would be where then line is tangent to the curve. However, here it looks like the extreme values will be at the "corners", (1,0), (0,1), (-1,0), and (0,1) where the curve does not have a tangent: answer 4.

after reading ur answer i think what they meant at the 1st question is the 2 pts from lagrange equations+4 points that they r the corners of the shape.
how could i knew how the shape look like?
i didnt understand ur answer to the last question - what do u mean:
"the extreme values will be where moving slightly one way or the other would take the OUT of the "box" formed by the closed curve"
i also didnt quite understood the rest of ur answer after that sentence

 

[mathwonk]

The answer of the opriginal question is obviously equal to -a, since you are maximizing the value of a linear function whose level surfaces are all planes parallel to the plane passing through the points (1,0,0), (0,1,0), and (0,0,1).

Visualizing this, you are minimizing the distance of such a plane from the origin while also passing through a point of the lower octant of the sphere of radius a. Since that octant is convex, it is nearest the origin at its corners, i.e. at (-a,0,0), and (0,-a),0) and (0,0,-a).