This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y, z) = x2 + y2 + z2; x4 + y4 + z4 = 13

The maximum value of f with the constraing x⁴+y⁴+z⁴ = 13 is 3 * √(13/3) = √39, and the minimum is √13

Step-by-step explanation:

Lets call g the restriction function. . The restriction we are given is g(x,y,z) = 13. The lagrange multiplier theorem says that if f reaches a maximum or minimum in a point (x₀,y₀,z₀) satisfying the constraing given by g, then we shoul have that for ceratin constant lambda.

This means that the extreme point should satisfy this conditions:

Now, lets compute each of the partial derivates, recall that when we compute the partial derivate of a function respect to a variable, lets say x, then we have to derivate that function treating the other variables like 'constants'. Thus, for example y² and z² from f will be treated as constants when we compute , thus, they will go away.

Therefore, the equations that we had before can be written this way:

Note that anyone of the first three equations will be solved as long as the variable involving it is equal to 0 (for example, if x = 0, then equation 1 is solved). We need a value different from 0 though, otherwise we wont be able to solve the last equation. Since all the first 3 equations are analogous, i will just solve the first one. The solutions will work on the second and third eqution as well, after changing x for y and z respectively.

2x = λ 4x³

(If x=0, it would be 0=0, therefore 0 is solution. I will assume x≠0)

1 = λ 4x²

x² = 1/(4λ)

x = √[ 1/(4λ) ];

or

x = -√[ 1/(4λ) ]

(or x=0)

Since λ is just a constant, then √[ 1/(4λ) ] is also a constant, we can call it c. We have then

x = c

x = -c

x = 0

as possible values of x.

y and z can take the same values.

We alredy know that the restriction wont apply if x=y=z=0. Lets suppose that 2 of x,y and z are 0. Then g(x,y,z) = c⁴ = 13, thus, the variable that is different from 0 should be equal to . In this case .

If only 1 of the variables is 0 and the others are c or -c, then g(x,y,z) = 2c⁴ = 13, or c⁴ = 6.5. Hence each variable should have the form . Here we will have

If none variables are 0, then x, y and z should be c or -c, and g(x,y,z) = 3c⁴ = 13. Thus, x, y and z should take a value of the form , and we therefore have

As a conclusion: the maximum value of f with the constraint x⁴+y⁴+z⁴ = 13 is 3 * √(13/3) = √39, and the minimum is √13.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

the answer is 62.8 cubic units