Mathematics, 31.03.2020 02:02, connersitte1221l

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

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answered: leasiamoore

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. g(x) = x^4+y^4+z^4 . 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 $\nabla{f}(x_0,y_0,z_0) = \lambda \, \nabla{g}(x_0,y_0,z_0)$ for ceratin constant lambda.

This means that the extreme point should satisfy this conditions:

$f_x(x_0,y_0,z_0) = \lambda \, g_x(x_0,y_0,z_0)$ $f_y(x_0,y_0,z_0) = \lambda \, g_y(x_0,y_0,z_0)$ $f_z(x_0,y_0,z_0) = \lamdba \, g_z(x_0,y_0,z_0)$ g(x_0,y_0,z_0) = 13

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 f_x , thus, they will go away.

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

$2x_0 = \lambda \, 4x_0^3$ $2y_0 = \lambda \, 4y_0^3$ $2z_0 = \lambda \, 4z_0^3$ x_0^4+y_0^4+z_0^4 = 13

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) ];

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 $^+_- \, ^4\sqrt {13}$ . In this case $f(x,y,z) = (^+_- \, ^4\sqrt {13})^2 = \sqrt{13} = 3.6056$ .

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 $^+_- \, ^4\sqrt{6.5}$ . Here we will have $f(x,y,z) = 2 * (^+_- \, ^4\sqrt{6.5})^2 = 2 * \sqrt{6.5} = 5.099$

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 $^+_- \, ^4\sqrt{13/3}$ , and we therefore have $f(x,y,z) = 3 * (^+_- \, ^4\sqrt{13/3})^2 = 3*\sqrt{13/3} = 6.245$

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.

answered: Guest

I’m pretty sure the slope would be 1 since the x and y values increase by 1
What is the slope of the line in the graph? 3+.2+-4-2-14 1 2 3 4 5 %2+-3 +